MATH 350 - DISCRETE
MATHEMATICS - Discrete random
variables and expected value
Question Bank - Set 5
Liberty University
Question 1
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 2) = 1
2, P (X= 3) = 1
4
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
x
x·P(X=x)
Step 2: In this case, we need to calculate E(X) using the given probability
mass function.
Step 3: First, we find the values of xand their corresponding probabilities:
For x=1:x·P(X= 1) = 1 ·1
4=1
4
For x=2:x·P(X= 2) = 2 ·1
2= 1
For x= 3 : x·P(X= 3) = 3 ·1
4=3
4
Step 4: Now, we sum up these values to find the expected value of X:
E(X) = 1
4+1+3
4=1
4+4
4+3
4=8
4= 2
Therefore, the expected value of the random variable Xis 2.
Question 2
Question
Let Xbe a discrete random variable with the following probability distribution:
X012
P(X) 0.2 0.5 0.3
Calculate the expected value of X.
Solution
To calculate the expected value of a discrete random variable, we use the for-
mula:
E(X) = X
all x
x·P(X=x)
Step 1: Calculate the expected value E(X) by multiplying each value of X
by its corresponding probability and summing the results.
E(X)=0·0.2+1·0.5+2·0.3
E(X) = 0 + 0.5+0.6
E(X)=1.1
Therefore, the expected value of the random variable Xis 1.1.
Question 3
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 1 3
P(X) 0.2 0.3 0.4 0.1
Find the expected value of X.
2
Solution
Step 1: The expected value of a discrete random variable Xwith probability
distribution P(X) is given by the formula:
E(X) = X
all x
x·P(X=x)
where xranges over all possible values of X.
Step 2: Calculate the expected value of Xusing the given probability dis-
tribution:
E(X)=(−2)(0.2) + (0)(0.3) + (1)(0.4) + (3)(0.1)
E(X) = −0.4+0+0.4+0.3
E(X)=0.3
Therefore, the expected value of the random variable Xis 0.3 .
Question 4
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) = (kx if x= 1,2,3
0 otherwise
Determine the value of kthat makes this a valid probability mass function,
and calculate the expected value of X.
Solution
Step 1: Determine the value of kto make this a valid probability mass function.
For a probability mass function, the sum of the probabilities over the entire
sample space should equal 1. Therefore, we must have:
X
x
P(X=x)=1
3
X
x=1
kx = 1
k(1) + k(2) + k(3) = 1
3
k+ 2k+ 3k= 1
6k= 1
k=1
6
Therefore, k=1
6.
Step 2: Calculate the expected value of X.
The expected value of a discrete random variable Xis given by:
E(X) = X
x
x·P(X=x)
E(X) =
3
X
x=1
x·1
6x
E(X) = 1
6(1) + 1
3(2) + 1
2(3)
E(X) = 1
6+2
6+3
6
E(X) = 6
6
E(X) = 1
Therefore, the expected value of Xis 1.
Question 5
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 1 4
P(X=x)k2k3k
Determine the value of ksuch that E(X) = 0, where E(X) denotes the
expected value of X.
4
Solution
Step 1: The expected value E(X) of a discrete random variable Xcan be
calculated using the formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values xthat Xcan take on.
Step 2: In this case, we have:
E(X) = (−2) ·k+ 1 ·2k+ 4 ·3k
Step 3: We are given that E(X) = 0, so we set the expression above equal
to 0 and solve for k:
−2k+ 2k+ 12k= 0
12k= 0
Step 4: Solving for k, we find:
k= 0
Therefore, the value of ksuch that E(X) = 0 is k= 0.
Question 6
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.2, P (X= 3) = 0.3, P (X= 5) = 0.4
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) into the formula:
E(X) = (−2) ·0.1 + (0) ·0.2 + (3) ·0.3 + (5) ·0.4
5
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.9+2
Step 4: Sum the values to find the expected value of X:
E(X)=2.7
Therefore, the expected value of the random variable Xis 2.7.
Question 7
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.1, P (X= 2) = 0.15, P (X= 3) = 0.3, P (X= 4) = 0.2, P (X= 5) = 0.25
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
all x
x·P(X=x)
Step 2: First, we calculate the expected value using the given probability
mass function:
E(X)=1×0.1+2×0.15 + 3 ×0.3+4×0.2+5×0.25
Step 3: Simplifying the expression:
E(X)=0.1+0.3+0.9+0.8+1.25
Step 4: Adding the terms together:
E(X)=3.35
Therefore, the expected value of the random variable Xis 3.35.
6
Question 8
Question
Let Xbe a discrete random variable with the following probability distribution:
X01234
P(X) 0.2 0.3 0.1 0.2 0.2
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by:
E(X) = X
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: Substituting the given values into the formula, we have:
E(X)=0·0.2+1·0.3+2·0.1+3·0.2+4·0.2.
Step 3: Calculating the expected value, we get:
E(X) = 0 + 0.3+0.2+0.6+0.8=2.1.
Therefore, the expected value of the random variable Xis 2.1.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
10, P (X= 0) = 3
10, P (X= 2) = 3
10, P (X= 4) = 3
10.
Find the expected value of X.
Solution
Step 1: Calculate the expected value using the formula E(X) = Pxi·P(X=
xi), where xiare the possible values of X.
Step 2: Substitute the values of xiand P(X=xi) into the formula.
Step 3: Calculate the expected value.
7
E(X)=(−2) ·1
10 + (0) ·3
10 + (2) ·3
10 + (4) ·3
10
E(X) = −2
10 +0+ 6
10 +12
10
E(X) = 16
10 = 1.6
Therefore, the expected value of the random variable Xis 1.6.
Question 10
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 0) = 1
4, P (X= 1) = 1
2, P (X= 2) = 1
8, P (X= 3) = 1
8
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by the formula:
E(X) = X
x
x·P(X=x)
Step 2: Calculate the expected value by plugging in the values of Xand
P(X) into the formula:
E(X) = (0)(1
4) + (1)(1
2) + (2)(1
8) + (3)(1
8)
Step 3: Simplify the expression:
E(X) = 0 + 1
2+2
8+3
8=1
2+5
8
Step 4: Find a common denominator and add the fractions:
E(X) = 4
8+5
8=9
8
Step 5: Therefore, the expected value of the random variable Xis 9
8.
8
Question 11
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=k) = (c
2kif k= 1,2,3,...,
0 otherwise,
Find the value of the constant cthat makes P(X=k) a valid probability
mass function. Then, calculate the expected value E[X].
Solution
Step 1: Find the value of cthat makes P(X=k) a valid probability mass
function.
Since P(X=k) is a probability mass function, we have:
∞
X
k=1
P(X=k) = 1
This gives us:
∞
X
k=1
c
2k= 1
c
∞
X
k=1
1
2k= 1
c1 + 1
2+1
22+. . .= 1
c1
1−1/2= 1
c×2=1
c=1
2
Therefore, the value of cthat makes P(X=k) a valid probability mass
function is 1
2.
Step 2: Calculate the expected value E[X].
The expected value of a random variable Xis given by:
E[X] =
∞
X
k=1
k·P(X=k)
9
E[X] =
∞
X
k=1
k·1
2k
E[X] =
∞
X
k=1
k
2k
To find the expected value, we can use the formula for the expected value of
a geometric random variable, which is 1−p
p2where pis the probability of success.
In this case, p=1
2, so the expected value E[X] is:
E[X] = 1−1
2
1
22=
1
2
1
4
= 2
Therefore, the expected value of the random variable Xis 2.
Question 12
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−1) = 0.3
P(X= 0) = 0.2
P(X= 1) = 0.5
Find the expected value of X.
Solution
Step 1: The expected value (or mean) of a discrete random variable Xis given
by the formula:
E(X) = Xx·P(X=x)
Step 2: Substitute the values of P(X=−1), P(X= 0), and P(X= 1) into
the formula:
E(X)=(−1) ·0.3 + (0) ·0.2 + (1) ·0.5
Step 3: Simplify the expression:
E(X) = −0.3+0+0.5
E(X)=0.2
Step 4: Therefore, the expected value of the random variable Xis 0.2.
10
Question 13
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 0) = 0.4, P (X= 1) = 0.3, P (X= 2) = 0.2
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by the formula:
E(X) = X
all x
x·P(X=x)
Step 2: Let’s calculate the expected value of Xusing the given probability
mass function:
E(X)=(−1) ·0.1 + (0) ·0.4 + (1) ·0.3 + (2) ·0.2
Step 3: Simplifying the expression, we get:
E(X) = −0.1+0+0.3+0.4=0.6
Step 4: Therefore, the expected value of the discrete random variable Xis
0.6 .
Question 14
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
2
9if x= 1
5
9if x= 2
2
9if x= 3
0 otherwise
Calculate the expected value of X.
11
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
all x
x·P(X=x)
Step 2: We can calculate the expected value of Xby using the formula above
and the given probability mass function:
E(X)=1·2
9+ 2 ·5
9+ 3 ·2
9
Step 3: Simplifying the expression:
E(X) = 2
9+10
9+6
9
Step 4: Combining the fractions:
E(X) = 18
9= 2
Therefore, the expected value of Xis 2.
Question 15
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 0) = 0.3, P (X= 1) = 0.4, P (X= 2) = 0.2
Determine the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 2: Substitute the values from the probability mass function into the
formula for expected value:
E(X)=(−1) ·0.1 + (0) ·0.3 + (1) ·0.4 + (2) ·0.2
12
Step 3: Simplify the expression:
E(X) = −0.1+0+0.4+0.4=0.7
Step 4: Therefore, the expected value of the random variable Xis 0.7.
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 3
P(X) 0.3 0.4 0.3
Find the expected value of X,E(X).
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by the formula:
E(X) = X
all x
x·P(X=x)
Step 2: Substitute the values of Xand P(X) from the probability distribu-
tion into the formula:
E(X)=(−2) ·0.3 + (0) ·0.4 + (3) ·0.3
Step 3: Calculate the expected value:
E(X) = −0.6+0+0.9
E(X)=0.3
Step 4: Therefore, the expected value of X,E(X), is 0.3 .
Question 17
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=k) = 1
2k, k = 1,2,3, . . .
Determine the expected value of X.
13
Solution
Step 1: First, we need to find the expected value of Xusing the formula:
E(X) =
∞
X
k=1
k·P(X=k)
Step 2: Substitute the given probability mass function into the formula:
E(X) =
∞
X
k=1
k·1
2k
Step 3: To simplify this expression, notice that this is a geometric series.
We can express the sum in a closed-form using the formula for the sum of an
infinite geometric series:
∞
X
k=1
ark−1=a
1−r,|r|<1
Step 4: By comparing with the formula above, we can see that a= 1 and
r=1
2. Thus, we have:
E(X) = 1
1−1
2
=1
1
2
= 2
Step 5: Therefore, the expected value of the random variable Xis 2 .
Question 18
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.5, P (X= 1) = 0.3, P (X= 2) = 0.1
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
as E(X) or µ, is calculated as:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) is the probability mass
function.
14
Step 2: Substitute the given values into the formula:
E(X) = (−2) ·0.1 + (0) ·0.5 + (1) ·0.3 + (2) ·0.1
Step 3: Perform the calculations:
E(X) = −0.2+0+0.3+0.2 = 0.3
Therefore, the expected value of random variable Xis 0.3 .
Question 19
Question
Let Xbe a discrete random variable with the following probability distribution:
X P (X=x)
0 0.1
1 0.3
2 0.2
3 0.4
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
all x
x·P(X=x)
Step 2: Let’s calculate the expected value E(X) using the provided proba-
bility distribution.
E(X)=0·0.1+1·0.3+2·0.2+3·0.4
Step 3: Calculate the values and multiply:
E(X) = 0 + 0.3+0.4+1.2
Step 4: Add the values together:
E(X)=1.9
Therefore, the expected value of Xis 1.9.
15
Question 20
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.3, P (X= 2) = 0.6.
Determine the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E[X] = X
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: In this case, we have three possible values for X: -2, 0, and 2.
Therefore, the expected value of Xis:
E[X]=(−2) ·P(X=−2) + (0) ·P(X= 0) + (2) ·P(X= 2).
Step 3: Substitute the given probabilities into the formula above:
E[X] = (−2) ·0.1 + (0) ·0.3 + (2) ·0.6.
Step 4: Calculate the expected value:
E[X] = −0.2+0+1.2=1.
Step 5: Therefore, the expected value of the random variable Xis 1 .
Question 21
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 1 5
P(X=x) 0.2 0.5 0.3
Find the expected value of X.
16
Solution
Step 1: Recall that the expected value (or mean) of a discrete random variable
Xis given by:
E(X) = X
all x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) from the given probability
distribution into the formula:
E(X)=(−2)(0.2) + (1)(0.5) + (5)(0.3)
Step 3: Calculate the expected value E(X):
E(X) = −0.4+0.5+1.5=1.6
Therefore, the expected value of the random variable Xis 1.6.
Question 22
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
3, P (X= 2) = 1
6, P (X= 3) = 1
2
Find the expected value E(X) of X.
Solution
To find the expected value E(X) of X, we use the formula:
E(X) = X
x
x·P(X=x)
Step 1: Find the expected value
E(X) = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
= 1 ·1
3+ 2 ·1
6+ 3 ·1
2
=1
3+1
3+3
2
=1+1+3
3
=5
3
Therefore, the expected value of Xis 5
3.
17
Question 23
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=k) = 1
2k,for k∈ {1,2,3, ...}
Find the expected value of X.
Solution
Step 1: To find the expected value of X, we use the formula:
E(X) =
∞
X
k=1
k·P(X=k)
Step 2: Substituting the given probability mass function P(X=k) = 1
2k
into the formula, we get:
E(X) =
∞
X
k=1
k·1
2k
Step 3: We can rewrite this sum to better analyze it:
E(X) = 1
2+2
4+3
8+4
16 +. . .
Step 4: We notice that this is a geometric series:
E(X) =
∞
X
k=1
k
2k=1
2+2
4+3
8+4
16 +. . .
Step 5: To find the sum of this series, we can multiply E(X) by 1
2and
subtract to get E(X)−1
2E(X):
1
2E(X) = 1
4+2
8+3
16 +4
32 +. . .
Step 6: Subtracting 1
2E(X) from E(X), we have:
1
2E(X) = 1
2+1
4+1
8+1
16 +. . .
Step 7: Recognizing that the right side of the equation is a geometric series
with first term 1
2and common ratio 1
2, we can use the formula for the sum of
an infinite geometric series:
1
2E(X) =
1
2
1−1
2
18
1
2E(X) = 1
Step 8: Solving for E(X):
E(X) = 2
Therefore, the expected value of Xis 2.
Question 24
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) = (1
2xfor x= 1,2,3, . . .
0 otherwise
Calculate the expected value of X.
Solution
Step 1: First, let’s recall the definition of expected value for a discrete random
variable X:
The expected value of X, denoted E(X), is calculated as:
E(X) = Xx·P(X=x)
Step 2: Using the given probability mass function, we can calculate the
expected value of Xas follows:
E(X) =
∞
X
x=1
x·1
2x
Step 3: To find this sum, we can note that it is a geometric series.
The sum of an infinite geometric series P∞
n=0 arnconverges to a
1−rif |r|<1.
In our case, a= 1 and r=1
2, so the expected value simplifies to:
E(X) = 1 ·1
1−1
2
= 1 ·2=2
Therefore, the expected value of the given discrete random variable Xis 2 .
19
Question 25
Question
Let X be a discrete random variable with the following probability mass function:
P(X= 1) = 1
4, P (X= 2) = 1
3, P (X= 3) = 1
6, P (X= 4) = 1
12
Calculate the expected value of X.
Solution
Step 1: To find the expected value of a discrete random variable X, we use the
formula:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of X and P(X=xi) are their respective
probabilities.
Step 2: Given that the values of X are 1, 2, 3, and 4 with probabilities 1
4,1
3,
1
6, and 1
12 respectively, we can calculate the expected value as:
E(X)=1·1
4+ 2 ·1
3+ 3 ·1
6+ 4 ·1
12
Step 3: Simplifying the expression, we get:
E(X) = 1
4+2
3+1
2+1
3=3
4+4
3=9
12 +16
12 =25
12
Therefore, the expected value of the random variable X is 25
12 .
Question 26
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−2) = 1
6, P (X= 1) = 1
3, P (X= 2) = 1
2
Calculate the expected value of X.
20
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the values from the probability mass function into the
formula:
E(X)=(−2) ·1
6+ (1) ·1
3+ (2) ·1
2
Step 3: Simplify the expression:
E(X) = −2
6+1
3+ 1
Step 4: Further simplify the expression:
E(X) = −1
3+1
3+ 1
Step 5: Therefore, the expected value of the random variable Xis:
E(X) = 1
Question 27
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 2) = 1
8, P (X= 3) = 1
2, P (X= 4) = 1
8.
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E[X] = X
x
x·P(X=x),
where xranges over all possible values of X.
Step 2: Compute the expected value of Xusing the given probability mass
function:
E[X]=1·1
4+ 2 ·1
8+ 3 ·1
2+ 4 ·1
8.
21
Step 3: Simplify the expression:
E[X] = 1
4+2
8+3
2+4
8.
Step 4: Combine the terms:
E[X] = 1
4+1
4+3
2
Step 5: Calculate the final result:
E[X] = 1
2+3
2= 2.
Therefore, the expected value of the random variable Xis 2.
Question 28
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.4, P (X= 1) = 0.3,and P(X=x)=0.2 for an unknown value x.
Calculate the expected value of X.
Solution
To find the expected value (mean) of a discrete random variable, we use the
formula:
E(X) = X
i
xi·P(X=xi)
where xiare all the possible values that Xcan take.
Step 1: Calculate the expected value of XWe will substitute the given
probability mass function values into the formula:
E(X) = (−2) ·0.1 + (0) ·0.4 + (1) ·0.3 + x·0.2
Simplifying, we get:
E(X) = −0.2+0+0.3+0.2x
E(X)=0.1+0.2x
So, the expected value of Xis 0.1+0.2x.
22
Question 29
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
2, P (X= 2) = 1
4, P (X= 3) = 1
8, P (X= 4) = 1
16, P (X= 100) = 1
16
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis defined as
E(X) = X
x
x·P(X=x)
Step 2: Substitute the given probabilities into the formula for expected value:
E(X) = 1 ·1
2+ 2 ·1
4+ 3 ·1
8+ 4 ·1
16 + 100 ·1
16
Step 3: Simplify the expression:
E(X) = 1
2+1
2+3
8+1
4+25
4
E(X) = 4
8+4
8+3
8+2
8+200
8
E(X) = 213
8
Therefore, the expected value of the random variable Xis 213
8.
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2
3 4
P(X) 0.3 0.2
0.4 0.1
Find the expected value of X.
23
Solution
To find the expected value of a discrete random variable X, we use the formula:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) are their corresponding
probabilities.
Step 1: Calculate the expected value using the formula.
E(X)=1·0.3+2·0.2+3·0.4+4·0.1
Step 2: Simplify the expression.
E(X)=0.3+0.4+1.2+0.4
E(X)=2.3
Therefore, the expected value of Xis 2.3.
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = (1
2kfor k= 1,2,3, . . .
0 otherwise
Find the expected value of X.
Solution
Step 1: First, we recall that the expected value of a discrete random variable X
is defined as E(X) = Pkk·P(X=k).
Step 2: We can now calculate the expected value of Xusing the given
probability mass function:
E(X) =
∞
X
k=1
k·P(X=k) =
∞
X
k=1
k·1
2k
Step 3: Let’s simplify the expression by expanding the summation:
E(X) = 1 ·1
2+ 2 ·1
4+ 3 ·1
8+. . .
24
Step 4: We can rewrite the summation in a more convenient form:
E(X) = 1
2+2
4+3
8+. . .
Step 5: Notice that the expression represents a geometric series with first
term a=1
2and common ratio r=1
2.
Step 6: The sum of an infinite geometric series is given by S=a
1−rfor
−1< r < 1.
Step 7: Applying the formula, we find the expected value of X:
E(X) =
1
2
1−1
2
=
1
2
1
2
= 1
Step 8: Therefore, the expected value of the discrete random variable Xis
1 .
Question 32
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X= 1) = 1
3, P (X= 2) = 1
2, P (X= 3) = 1
6
Calculate the expected value of X,E(X).
Solution
Step 1: Recall that the expected value of a discrete random variable Xwith
probability mass function p(x) is given by:
E(X) = X
all x
x·p(x)
Step 2: Calculate the expected value of Xusing the given probability mass
function:
E(X) = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
E(X) = 1 ·1
3+ 2 ·1
2+ 3 ·1
6
E(X) = 1
3+1+1
2
Step 3: Simplify the expression:
E(X) = 1
3+2
2+1
2
25
E(X) = 1+4+3
6
E(X) = 8
6
Step 4: Further simplify to get the final answer:
E(X) = 4
3
Therefore, the expected value of Xis 4
3.
Question 33
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
3
10 if x= 1
1
5if x= 2
1
10 if x= 3
2
5if x= 4
0 otherwise
Calculate the expected value of X.
Solution
Step 1: First, recall that the expected value of a discrete random variable X
with probability mass function P(X=x) is given by:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the values from the probability mass function into the
formula for expected value:
E(X)=1·3
10 + 2 ·1
5+ 3 ·1
10 + 4 ·2
5
Step 3: Simplify the expression:
E(X) = 3
10 +2
5+3
10 +8
5
E(X) = 3+4+3+16
10
26
E(X) = 26
10
E(X)=2.6
Therefore, the expected value of the random variable Xis 2.6.
Question 34
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = (2
3kif k≥1
0 otherwise
Determine the expected value of X.
Solution
Step 1: First, we need to calculate the expected value of Xusing the definition
of expected value for a discrete random variable. The expected value of X,
denoted by E(X), is given by:
E(X) = X
k
k·P(X=k)
Step 2: Substituting the given probability mass function into the formula
for the expected value:
E(X) =
∞
X
k=1
k·2
3k
Step 3: The above expression can be rewritten as:
E(X) =
∞
X
k=1
2k
3k
Step 4: To simplify the summation, we can differentiate the power series
representation of a function. Let f(x) = P∞
k=0 xk. Then, we have:
f′(x) =
∞
X
k=1
kxk−1=
∞
X
k=0
kxk
Step 5: Now, differentiating f(x) = P∞
k=0 xkterm by term gives:
f′(x) = d
dx ∞
X
k=0
xk!=
∞
X
k=0
d
dx(xk) =
∞
X
k=1
kxk−1
27
Step 6: Since f(x) = P∞
k=0 xk=1
1−xfor |x|<1, we have:
f′(x) = d
dx 1
1−x=1
(1 −x)2
Step 7: Comparing the above results with E(X), we see that E(X) =
f′(1/3). Using the result obtained for f′(x):
E(X) = 1
(1 −1
3)2=1
(2
3)2= 9
Therefore, the expected value of the random variable Xis 9.
Question 35
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
4, P (X= 0) = 1
2, P (X= 5) = 1
4.
Calculate the expected value of X.
Solution
Step 1: First, we recall that the expected value (or mean) of a discrete random
variable Xis given by:
E(X) = X
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: In this case, we can calculate the expected value E(X) as follows:
E(X)=(−2) ·1
4+ (0) ·1
2+ (5) ·1
4
E(X) = −1
2+0+5
4
E(X) = 3
4.
Therefore, the expected value of the random variable Xis 3
4.
28
Step 4: Now, we sum up these values to find the expected value of X:
E(X) = 1
4+1+3
4=1
4+4
4+3
4=8
4= 2
Therefore, the expected value of the random variable Xis 2.
Question 2
Question
Let Xbe a discrete random variable with the following probability distribution:
X012
P(X) 0.2 0.5 0.3
Calculate the expected value of X.
Solution
To calculate the expected value of a discrete random variable, we use the for-
mula:
E(X) = X
all x
x·P(X=x)
Step 1: Calculate the expected value E(X) by multiplying each value of X
by its corresponding probability and summing the results.
E(X)=0·0.2+1·0.5+2·0.3
E(X) = 0 + 0.5+0.6
E(X)=1.1
Therefore, the expected value of the random variable Xis 1.1.
Question 3
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 1 3
P(X) 0.2 0.3 0.4 0.1
Find the expected value of X.
2
Solution
Step 1: The expected value of a discrete random variable Xwith probability
distribution P(X) is given by the formula:
E(X) = X
all x
x·P(X=x)
where xranges over all possible values of X.
Step 2: Calculate the expected value of Xusing the given probability dis-
tribution:
E(X)=(−2)(0.2) + (0)(0.3) + (1)(0.4) + (3)(0.1)
E(X) = −0.4+0+0.4+0.3
E(X)=0.3
Therefore, the expected value of the random variable Xis 0.3 .
Question 4
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) = (kx if x= 1,2,3
0 otherwise
Determine the value of kthat makes this a valid probability mass function,
and calculate the expected value of X.
Solution
Step 1: Determine the value of kto make this a valid probability mass function.
For a probability mass function, the sum of the probabilities over the entire
sample space should equal 1. Therefore, we must have:
X
x
P(X=x)=1
3
X
x=1
kx = 1
k(1) + k(2) + k(3) = 1
3
k+ 2k+ 3k= 1
6k= 1
k=1
6
Therefore, k=1
6.
Step 2: Calculate the expected value of X.
The expected value of a discrete random variable Xis given by:
E(X) = X
x
x·P(X=x)
E(X) =
3
X
x=1
x·1
6x
E(X) = 1
6(1) + 1
3(2) + 1
2(3)
E(X) = 1
6+2
6+3
6
E(X) = 6
6
E(X) = 1
Therefore, the expected value of Xis 1.
Question 5
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 1 4
P(X=x)k2k3k
Determine the value of ksuch that E(X) = 0, where E(X) denotes the
expected value of X.
4
Solution
Step 1: The expected value E(X) of a discrete random variable Xcan be
calculated using the formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values xthat Xcan take on.
Step 2: In this case, we have:
E(X) = (−2) ·k+ 1 ·2k+ 4 ·3k
Step 3: We are given that E(X) = 0, so we set the expression above equal
to 0 and solve for k:
−2k+ 2k+ 12k= 0
12k= 0
Step 4: Solving for k, we find:
k= 0
Therefore, the value of ksuch that E(X) = 0 is k= 0.
Question 6
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.2, P (X= 3) = 0.3, P (X= 5) = 0.4
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) into the formula:
E(X) = (−2) ·0.1 + (0) ·0.2 + (3) ·0.3 + (5) ·0.4
5
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.9+2
Step 4: Sum the values to find the expected value of X:
E(X)=2.7
Therefore, the expected value of the random variable Xis 2.7.
Question 7
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.1, P (X= 2) = 0.15, P (X= 3) = 0.3, P (X= 4) = 0.2, P (X= 5) = 0.25
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
all x
x·P(X=x)
Step 2: First, we calculate the expected value using the given probability
mass function:
E(X)=1×0.1+2×0.15 + 3 ×0.3+4×0.2+5×0.25
Step 3: Simplifying the expression:
E(X)=0.1+0.3+0.9+0.8+1.25
Step 4: Adding the terms together:
E(X)=3.35
Therefore, the expected value of the random variable Xis 3.35.
6
Question 8
Question
Let Xbe a discrete random variable with the following probability distribution:
X01234
P(X) 0.2 0.3 0.1 0.2 0.2
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by:
E(X) = X
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: Substituting the given values into the formula, we have:
E(X)=0·0.2+1·0.3+2·0.1+3·0.2+4·0.2.
Step 3: Calculating the expected value, we get:
E(X) = 0 + 0.3+0.2+0.6+0.8=2.1.
Therefore, the expected value of the random variable Xis 2.1.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
10, P (X= 0) = 3
10, P (X= 2) = 3
10, P (X= 4) = 3
10.
Find the expected value of X.
Solution
Step 1: Calculate the expected value using the formula E(X) = Pxi·P(X=
xi), where xiare the possible values of X.
Step 2: Substitute the values of xiand P(X=xi) into the formula.
Step 3: Calculate the expected value.
7
E(X)=(−2) ·1
10 + (0) ·3
10 + (2) ·3
10 + (4) ·3
10
E(X) = −2
10 +0+ 6
10 +12
10
E(X) = 16
10 = 1.6
Therefore, the expected value of the random variable Xis 1.6.
Question 10
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 0) = 1
4, P (X= 1) = 1
2, P (X= 2) = 1
8, P (X= 3) = 1
8
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by the formula:
E(X) = X
x
x·P(X=x)
Step 2: Calculate the expected value by plugging in the values of Xand
P(X) into the formula:
E(X) = (0)(1
4) + (1)(1
2) + (2)(1
8) + (3)(1
8)
Step 3: Simplify the expression:
E(X) = 0 + 1
2+2
8+3
8=1
2+5
8
Step 4: Find a common denominator and add the fractions:
E(X) = 4
8+5
8=9
8
Step 5: Therefore, the expected value of the random variable Xis 9
8.
8
Question 11
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=k) = (c
2kif k= 1,2,3,...,
0 otherwise,
Find the value of the constant cthat makes P(X=k) a valid probability
mass function. Then, calculate the expected value E[X].
Solution
Step 1: Find the value of cthat makes P(X=k) a valid probability mass
function.
Since P(X=k) is a probability mass function, we have:
∞
X
k=1
P(X=k) = 1
This gives us:
∞
X
k=1
c
2k= 1
c
∞
X
k=1
1
2k= 1
c1 + 1
2+1
22+. . .= 1
c1
1−1/2= 1
c×2=1
c=1
2
Therefore, the value of cthat makes P(X=k) a valid probability mass
function is 1
2.
Step 2: Calculate the expected value E[X].
The expected value of a random variable Xis given by:
E[X] =
∞
X
k=1
k·P(X=k)
9
E[X] =
∞
X
k=1
k·1
2k
E[X] =
∞
X
k=1
k
2k
To find the expected value, we can use the formula for the expected value of
a geometric random variable, which is 1−p
p2where pis the probability of success.
In this case, p=1
2, so the expected value E[X] is:
E[X] = 1−1
2
1
22=
1
2
1
4
= 2
Therefore, the expected value of the random variable Xis 2.
Question 12
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−1) = 0.3
P(X= 0) = 0.2
P(X= 1) = 0.5
Find the expected value of X.
Solution
Step 1: The expected value (or mean) of a discrete random variable Xis given
by the formula:
E(X) = Xx·P(X=x)
Step 2: Substitute the values of P(X=−1), P(X= 0), and P(X= 1) into
the formula:
E(X)=(−1) ·0.3 + (0) ·0.2 + (1) ·0.5
Step 3: Simplify the expression:
E(X) = −0.3+0+0.5
E(X)=0.2
Step 4: Therefore, the expected value of the random variable Xis 0.2.
10
Question 13
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 0) = 0.4, P (X= 1) = 0.3, P (X= 2) = 0.2
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by the formula:
E(X) = X
all x
x·P(X=x)
Step 2: Let’s calculate the expected value of Xusing the given probability
mass function:
E(X)=(−1) ·0.1 + (0) ·0.4 + (1) ·0.3 + (2) ·0.2
Step 3: Simplifying the expression, we get:
E(X) = −0.1+0+0.3+0.4=0.6
Step 4: Therefore, the expected value of the discrete random variable Xis
0.6 .
Question 14
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
2
9if x= 1
5
9if x= 2
2
9if x= 3
0 otherwise
Calculate the expected value of X.
11
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
all x
x·P(X=x)
Step 2: We can calculate the expected value of Xby using the formula above
and the given probability mass function:
E(X)=1·2
9+ 2 ·5
9+ 3 ·2
9
Step 3: Simplifying the expression:
E(X) = 2
9+10
9+6
9
Step 4: Combining the fractions:
E(X) = 18
9= 2
Therefore, the expected value of Xis 2.
Question 15
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 0) = 0.3, P (X= 1) = 0.4, P (X= 2) = 0.2
Determine the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 2: Substitute the values from the probability mass function into the
formula for expected value:
E(X)=(−1) ·0.1 + (0) ·0.3 + (1) ·0.4 + (2) ·0.2
12
Step 3: Simplify the expression:
E(X) = −0.1+0+0.4+0.4=0.7
Step 4: Therefore, the expected value of the random variable Xis 0.7.
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 3
P(X) 0.3 0.4 0.3
Find the expected value of X,E(X).
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by the formula:
E(X) = X
all x
x·P(X=x)
Step 2: Substitute the values of Xand P(X) from the probability distribu-
tion into the formula:
E(X)=(−2) ·0.3 + (0) ·0.4 + (3) ·0.3
Step 3: Calculate the expected value:
E(X) = −0.6+0+0.9
E(X)=0.3
Step 4: Therefore, the expected value of X,E(X), is 0.3 .
Question 17
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=k) = 1
2k, k = 1,2,3, . . .
Determine the expected value of X.
13
Solution
Step 1: First, we need to find the expected value of Xusing the formula:
E(X) =
∞
X
k=1
k·P(X=k)
Step 2: Substitute the given probability mass function into the formula:
E(X) =
∞
X
k=1
k·1
2k
Step 3: To simplify this expression, notice that this is a geometric series.
We can express the sum in a closed-form using the formula for the sum of an
infinite geometric series:
∞
X
k=1
ark−1=a
1−r,|r|<1
Step 4: By comparing with the formula above, we can see that a= 1 and
r=1
2. Thus, we have:
E(X) = 1
1−1
2
=1
1
2
= 2
Step 5: Therefore, the expected value of the random variable Xis 2 .
Question 18
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.5, P (X= 1) = 0.3, P (X= 2) = 0.1
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
as E(X) or µ, is calculated as:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) is the probability mass
function.
14
Step 2: Substitute the given values into the formula:
E(X) = (−2) ·0.1 + (0) ·0.5 + (1) ·0.3 + (2) ·0.1
Step 3: Perform the calculations:
E(X) = −0.2+0+0.3+0.2 = 0.3
Therefore, the expected value of random variable Xis 0.3 .
Question 19
Question
Let Xbe a discrete random variable with the following probability distribution:
X P (X=x)
0 0.1
1 0.3
2 0.2
3 0.4
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis given by the
formula:
E(X) = X
all x
x·P(X=x)
Step 2: Let’s calculate the expected value E(X) using the provided proba-
bility distribution.
E(X)=0·0.1+1·0.3+2·0.2+3·0.4
Step 3: Calculate the values and multiply:
E(X) = 0 + 0.3+0.4+1.2
Step 4: Add the values together:
E(X)=1.9
Therefore, the expected value of Xis 1.9.
15
Question 20
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.3, P (X= 2) = 0.6.
Determine the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E[X] = X
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: In this case, we have three possible values for X: -2, 0, and 2.
Therefore, the expected value of Xis:
E[X]=(−2) ·P(X=−2) + (0) ·P(X= 0) + (2) ·P(X= 2).
Step 3: Substitute the given probabilities into the formula above:
E[X] = (−2) ·0.1 + (0) ·0.3 + (2) ·0.6.
Step 4: Calculate the expected value:
E[X] = −0.2+0+1.2=1.
Step 5: Therefore, the expected value of the random variable Xis 1 .
Question 21
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 1 5
P(X=x) 0.2 0.5 0.3
Find the expected value of X.
16
Solution
Step 1: Recall that the expected value (or mean) of a discrete random variable
Xis given by:
E(X) = X
all x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) from the given probability
distribution into the formula:
E(X)=(−2)(0.2) + (1)(0.5) + (5)(0.3)
Step 3: Calculate the expected value E(X):
E(X) = −0.4+0.5+1.5=1.6
Therefore, the expected value of the random variable Xis 1.6.
Question 22
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
3, P (X= 2) = 1
6, P (X= 3) = 1
2
Find the expected value E(X) of X.
Solution
To find the expected value E(X) of X, we use the formula:
E(X) = X
x
x·P(X=x)
Step 1: Find the expected value
E(X) = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
= 1 ·1
3+ 2 ·1
6+ 3 ·1
2
=1
3+1
3+3
2
=1+1+3
3
=5
3
Therefore, the expected value of Xis 5
3.
17
Question 23
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=k) = 1
2k,for k∈ {1,2,3, ...}
Find the expected value of X.
Solution
Step 1: To find the expected value of X, we use the formula:
E(X) =
∞
X
k=1
k·P(X=k)
Step 2: Substituting the given probability mass function P(X=k) = 1
2k
into the formula, we get:
E(X) =
∞
X
k=1
k·1
2k
Step 3: We can rewrite this sum to better analyze it:
E(X) = 1
2+2
4+3
8+4
16 +. . .
Step 4: We notice that this is a geometric series:
E(X) =
∞
X
k=1
k
2k=1
2+2
4+3
8+4
16 +. . .
Step 5: To find the sum of this series, we can multiply E(X) by 1
2and
subtract to get E(X)−1
2E(X):
1
2E(X) = 1
4+2
8+3
16 +4
32 +. . .
Step 6: Subtracting 1
2E(X) from E(X), we have:
1
2E(X) = 1
2+1
4+1
8+1
16 +. . .
Step 7: Recognizing that the right side of the equation is a geometric series
with first term 1
2and common ratio 1
2, we can use the formula for the sum of
an infinite geometric series:
1
2E(X) =
1
2
1−1
2
18
1
2E(X) = 1
Step 8: Solving for E(X):
E(X) = 2
Therefore, the expected value of Xis 2.
Question 24
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) = (1
2xfor x= 1,2,3, . . .
0 otherwise
Calculate the expected value of X.
Solution
Step 1: First, let’s recall the definition of expected value for a discrete random
variable X:
The expected value of X, denoted E(X), is calculated as:
E(X) = Xx·P(X=x)
Step 2: Using the given probability mass function, we can calculate the
expected value of Xas follows:
E(X) =
∞
X
x=1
x·1
2x
Step 3: To find this sum, we can note that it is a geometric series.
The sum of an infinite geometric series P∞
n=0 arnconverges to a
1−rif |r|<1.
In our case, a= 1 and r=1
2, so the expected value simplifies to:
E(X) = 1 ·1
1−1
2
= 1 ·2=2
Therefore, the expected value of the given discrete random variable Xis 2 .
19
Question 25
Question
Let X be a discrete random variable with the following probability mass function:
P(X= 1) = 1
4, P (X= 2) = 1
3, P (X= 3) = 1
6, P (X= 4) = 1
12
Calculate the expected value of X.
Solution
Step 1: To find the expected value of a discrete random variable X, we use the
formula:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of X and P(X=xi) are their respective
probabilities.
Step 2: Given that the values of X are 1, 2, 3, and 4 with probabilities 1
4,1
3,
1
6, and 1
12 respectively, we can calculate the expected value as:
E(X)=1·1
4+ 2 ·1
3+ 3 ·1
6+ 4 ·1
12
Step 3: Simplifying the expression, we get:
E(X) = 1
4+2
3+1
2+1
3=3
4+4
3=9
12 +16
12 =25
12
Therefore, the expected value of the random variable X is 25
12 .
Question 26
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−2) = 1
6, P (X= 1) = 1
3, P (X= 2) = 1
2
Calculate the expected value of X.
20
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the values from the probability mass function into the
formula:
E(X)=(−2) ·1
6+ (1) ·1
3+ (2) ·1
2
Step 3: Simplify the expression:
E(X) = −2
6+1
3+ 1
Step 4: Further simplify the expression:
E(X) = −1
3+1
3+ 1
Step 5: Therefore, the expected value of the random variable Xis:
E(X) = 1
Question 27
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 2) = 1
8, P (X= 3) = 1
2, P (X= 4) = 1
8.
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E[X] = X
x
x·P(X=x),
where xranges over all possible values of X.
Step 2: Compute the expected value of Xusing the given probability mass
function:
E[X]=1·1
4+ 2 ·1
8+ 3 ·1
2+ 4 ·1
8.
21
Step 3: Simplify the expression:
E[X] = 1
4+2
8+3
2+4
8.
Step 4: Combine the terms:
E[X] = 1
4+1
4+3
2
Step 5: Calculate the final result:
E[X] = 1
2+3
2= 2.
Therefore, the expected value of the random variable Xis 2.
Question 28
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.4, P (X= 1) = 0.3,and P(X=x)=0.2 for an unknown value x.
Calculate the expected value of X.
Solution
To find the expected value (mean) of a discrete random variable, we use the
formula:
E(X) = X
i
xi·P(X=xi)
where xiare all the possible values that Xcan take.
Step 1: Calculate the expected value of XWe will substitute the given
probability mass function values into the formula:
E(X) = (−2) ·0.1 + (0) ·0.4 + (1) ·0.3 + x·0.2
Simplifying, we get:
E(X) = −0.2+0+0.3+0.2x
E(X)=0.1+0.2x
So, the expected value of Xis 0.1+0.2x.
22
Question 29
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
2, P (X= 2) = 1
4, P (X= 3) = 1
8, P (X= 4) = 1
16, P (X= 100) = 1
16
Calculate the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis defined as
E(X) = X
x
x·P(X=x)
Step 2: Substitute the given probabilities into the formula for expected value:
E(X) = 1 ·1
2+ 2 ·1
4+ 3 ·1
8+ 4 ·1
16 + 100 ·1
16
Step 3: Simplify the expression:
E(X) = 1
2+1
2+3
8+1
4+25
4
E(X) = 4
8+4
8+3
8+2
8+200
8
E(X) = 213
8
Therefore, the expected value of the random variable Xis 213
8.
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2
3 4
P(X) 0.3 0.2
0.4 0.1
Find the expected value of X.
23
Solution
To find the expected value of a discrete random variable X, we use the formula:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) are their corresponding
probabilities.
Step 1: Calculate the expected value using the formula.
E(X)=1·0.3+2·0.2+3·0.4+4·0.1
Step 2: Simplify the expression.
E(X)=0.3+0.4+1.2+0.4
E(X)=2.3
Therefore, the expected value of Xis 2.3.
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = (1
2kfor k= 1,2,3, . . .
0 otherwise
Find the expected value of X.
Solution
Step 1: First, we recall that the expected value of a discrete random variable X
is defined as E(X) = Pkk·P(X=k).
Step 2: We can now calculate the expected value of Xusing the given
probability mass function:
E(X) =
∞
X
k=1
k·P(X=k) =
∞
X
k=1
k·1
2k
Step 3: Let’s simplify the expression by expanding the summation:
E(X) = 1 ·1
2+ 2 ·1
4+ 3 ·1
8+. . .
24
Step 4: We can rewrite the summation in a more convenient form:
E(X) = 1
2+2
4+3
8+. . .
Step 5: Notice that the expression represents a geometric series with first
term a=1
2and common ratio r=1
2.
Step 6: The sum of an infinite geometric series is given by S=a
1−rfor
−1< r < 1.
Step 7: Applying the formula, we find the expected value of X:
E(X) =
1
2
1−1
2
=
1
2
1
2
= 1
Step 8: Therefore, the expected value of the discrete random variable Xis
1 .
Question 32
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X= 1) = 1
3, P (X= 2) = 1
2, P (X= 3) = 1
6
Calculate the expected value of X,E(X).
Solution
Step 1: Recall that the expected value of a discrete random variable Xwith
probability mass function p(x) is given by:
E(X) = X
all x
x·p(x)
Step 2: Calculate the expected value of Xusing the given probability mass
function:
E(X) = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
E(X) = 1 ·1
3+ 2 ·1
2+ 3 ·1
6
E(X) = 1
3+1+1
2
Step 3: Simplify the expression:
E(X) = 1
3+2
2+1
2
25
E(X) = 1+4+3
6
E(X) = 8
6
Step 4: Further simplify to get the final answer:
E(X) = 4
3
Therefore, the expected value of Xis 4
3.
Question 33
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
3
10 if x= 1
1
5if x= 2
1
10 if x= 3
2
5if x= 4
0 otherwise
Calculate the expected value of X.
Solution
Step 1: First, recall that the expected value of a discrete random variable X
with probability mass function P(X=x) is given by:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the values from the probability mass function into the
formula for expected value:
E(X)=1·3
10 + 2 ·1
5+ 3 ·1
10 + 4 ·2
5
Step 3: Simplify the expression:
E(X) = 3
10 +2
5+3
10 +8
5
E(X) = 3+4+3+16
10
26
E(X) = 26
10
E(X)=2.6
Therefore, the expected value of the random variable Xis 2.6.
Question 34
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = (2
3kif k≥1
0 otherwise
Determine the expected value of X.
Solution
Step 1: First, we need to calculate the expected value of Xusing the definition
of expected value for a discrete random variable. The expected value of X,
denoted by E(X), is given by:
E(X) = X
k
k·P(X=k)
Step 2: Substituting the given probability mass function into the formula
for the expected value:
E(X) =
∞
X
k=1
k·2
3k
Step 3: The above expression can be rewritten as:
E(X) =
∞
X
k=1
2k
3k
Step 4: To simplify the summation, we can differentiate the power series
representation of a function. Let f(x) = P∞
k=0 xk. Then, we have:
f′(x) =
∞
X
k=1
kxk−1=
∞
X
k=0
kxk
Step 5: Now, differentiating f(x) = P∞
k=0 xkterm by term gives:
f′(x) = d
dx ∞
X
k=0
xk!=
∞
X
k=0
d
dx(xk) =
∞
X
k=1
kxk−1
27
Step 6: Since f(x) = P∞
k=0 xk=1
1−xfor |x|<1, we have:
f′(x) = d
dx 1
1−x=1
(1 −x)2
Step 7: Comparing the above results with E(X), we see that E(X) =
f′(1/3). Using the result obtained for f′(x):
E(X) = 1
(1 −1
3)2=1
(2
3)2= 9
Therefore, the expected value of the random variable Xis 9.
Question 35
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
4, P (X= 0) = 1
2, P (X= 5) = 1
4.
Calculate the expected value of X.
Solution
Step 1: First, we recall that the expected value (or mean) of a discrete random
variable Xis given by:
E(X) = X
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: In this case, we can calculate the expected value E(X) as follows:
E(X)=(−2) ·1
4+ (0) ·1
2+ (5) ·1
4
E(X) = −1
2+0+5
4
E(X) = 3
4.
Therefore, the expected value of the random variable Xis 3
4.
28
