MATH 350 - DISCRETE
MATHEMATICS - Discrete random
variables and expected value
Question Bank - Set 1
Liberty University
Question 1
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
6, P (X= 2) = 1
3, P (X= 3) = 1
2
Find 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: Using the given probability mass function, we can calculate the
expected value of Xas follows:
E[X] = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
Step 3: Substitute the probabilities into the expression:
E[X] = 1 ·1
6+ 2 ·1
3+ 3 ·1
2
Step 4: Simplify the expression:
E[X] = 1
6+2
3+3
2
Step 5: Find a common denominator to add the fractions:
E[X] = 1
6+4
6+9
6
Step 6: Add the fractions together:
E[X] = 14
6=7
3
Therefore, the expected value of random variable Xis 7
3.
Question 2
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
3, P (X= 0) = 1
6, P (X= 1) = 1
2
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
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) are their corresponding
probabilities.
Step 2: Substitute the values of xiand P(X=xi) into the formula:
E(X)=(−1) ·1
3+ (0) ·1
6+ (1) ·1
2
Step 3: Simplify the expression:
E(X) = −1
3+0+1
2=1
2−1
3=3
6−2
6=1
6
Step 4: Therefore, the expected value of the random variable Xis 1
6.
2
Question 3
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=x) =
0.2 if x= 0
0.3 if x= 1
0.1 if x= 2
0.4 if x= 3
Determine the expected value of X.
Solution
Step 1: The expected value E(X) of a discrete random variable Xis defined as:
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)=0·0.2+1·0.3+2·0.1+3·0.4
Step 3: Simplifying the expression:
E(X) = 0 + 0.3+0.2+1.2
Step 4: Thus, the expected value of Xis:
E(X) = 1.7
Therefore, the expected value of Xis 1.7.
Question 4
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 4
P(X)1
6
1
2
1
3
Calculate the expected value of X.
3
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of X.
Step 2: Using the given probability distribution of X, we can calculate the
expected value as:
E(X)=(−2) ·1
6+ (0) ·1
2+ (4) ·1
3
Step 3: Simplifying the expression, we get:
E(X) = −2
6+0+4
3
E(X) = −1
3+4
3
E(X) = 3
3= 1
Therefore, the expected value of the random variable Xis 1.
Question 5
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
6, P (X= 0) = 1
3, P (X= 2) = 1
2.
Calculate the expected value of X.
Solution
Step 1: First, 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 given probability mass function into
the formula:
E(X)=(−2) ·1
6+ (0) ·1
3+ (2) ·1
2.
Step 3: Simplify the expression:
E(X) = −1
3+ 0 + 1 = 2
3.
Step 4: Therefore, the expected value of the random variable Xis 2
3.
4
Question 6
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2
P(X=x) 0.3 0.5 0.2
Find the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis defined as:
E(X) = X
all x
x·P(X=x)
Step 2: Using the given probability distribution, we can calculate the ex-
pected value of Xas:
E(X)=0·0.3+1·0.5+2·0.2
Step 3: Simplifying the expression, we get:
E(X) = 0 + 0.5+0.4
E(X)=0.9
Therefore, the expected value of the random variable Xis 0.9.
Question 7
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
8, P (X= 0) = 3
8, P (X= 2) = 1
4, P (X= 4) = 1
4.
Calculate the expected value of X.
5
Solution
Step 1: To find the expected value of a discrete random variable X, we use the
formula:
E(X) = X
all x
x·P(X=x).
Step 2: Given the probability mass function of X, we can substitute in the
values of xand P(X=x):
E(X)=(−2) ·1
8+ (0) ·3
8+ (2) ·1
4+ (4) ·1
4.
Step 3: Simplifying the expression:
E(X) = −2
8+0+2
4+4
4=−1
4+1
2+ 1.
Step 4: Further simplifying:
E(X) = 1
4+2
4+4
4=7
4.
Therefore, the expected value of the random variable Xis 7
4.
Question 8
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=x) = (1
2x+1 ,if x= 0,1,2, . . .
0,otherwise
Calculate the expected value of X.
Solution
Step 1: To find the expected value of a discrete random variable X, denoted by
E[X], we use the formula:
E[X] = X
x
x·P(X=x)
Step 2: Given the probability mass function of X, we need to calculate E[X]
as follows:
E[X] =
∞
X
x=0
x·1
2x+1
6
Step 3: Let’s simplify the expression before calculating the sum. Notice that
x·1
2x+1 can be rewritten as x
2x. So we have:
E[X] =
∞
X
x=0
x
2x
Step 4: To calculate this sum, we can differentiate a known power series.
Let’s consider the power series expansion of the function f(t) = P∞
x=0 tx. This
is a geometric series that converges to 1
1−tfor |t|<1.
Step 5: Differentiating both sides with respect to t, we get:
f′(t) =
∞
X
x=0
x·tx−1=d
dt 1
1−t
Step 6: Simplifying the right side and using the chain rule, we have:
∞
X
x=0
x·tx−1=1
(1 −t)2
Step 7: Setting t=1
2, we find:
∞
X
x=0
x·1
2x−1
=1
1−1
22= 4
Step 8: Therefore, the expected value of the random variable Xis E[X] = 4.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−3) = 0.1, P (X= 0) = 0.3, P (X= 2) = 0.4, P (X= 5) = 0.2
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)
where the sum is taken over all possible values of X.
7
Step 2: In this case, the possible values of Xare -3, 0, 2, and 5.
Step 3: We can now calculate the expected value:
E(X)=(−3)(0.1) + (0)(0.3) + (2)(0.4) + (5)(0.2)
Step 4: Simplifying, we get:
E(X) = −0.3+0+0.8 + 1 = 1.5
Step 5: Therefore, the expected value of the random variable Xis 1.5.
Question 10
Question
Let Xbe a discrete random variable with the following probability distribution:
x P (X=x)
1 0.2
2 0.3
3 0.1
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 the formula:
E(X) = Xx·P(X=x)
Step 2: We will substitute the values from the probability distribution of X
into the formula to calculate the expected value:
E(X)=1·0.2+2·0.3+3·0.1+5·0.4
Step 3: Calculate the expected value:
E(X) = 0.2+0.6+0.3 + 2 = 3.1
Therefore, the expected value of the random variable Xis 3.1.
Question 11
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
8
P(X=−2) = 1
9, P (X= 0) = 1
3, P (X= 1) = 1
6, P (X= 3) = 1
4
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
i
xi·P(X=xi)
where the sum is taken over all possible values xithat Xcan take.
Step 2: Substitute the values of Xand their corresponding probabilities into
the formula:
E(X) = (−2) 1
9+ (0) 1
3+ (1) 1
6+ (3) 1
4
Step 3: Simplify the expression by multiplying and adding the terms:
E(X) = −2
9+0+1
6+3
4
E(X) = −8
36 +0+ 6
36 +27
36
E(X) = 25
36
Step 4: Therefore, the expected value of the random variable Xis 25
36 .
Question 12
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
4, P (X= 0) = 1
2, P (X= 1) = 1
4
Calculate the expected value of X.
9
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: Substitute the given probabilities into the formula.
E(X)=(−1) ·1
4+ (0) ·1
2+ (1) ·1
4
Step 3: Simplify the expression.
E(X) = −1
4+0+1
4= 0
Step 4: Therefore, the expected value of the random variable Xis 0 .
Question 13
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−1) = 1
4, P (X= 0) = 1
2, P (X= 1) = 1
4
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis defined
as:
E[X] = X
all x
x·P(X=x)
Step 2: Using the given probability mass function for X, we have:
E[X]=(−1) ·1
4+ (0) ·1
2+ (1) ·1
4
Step 3: Simplifying the expression, we get:
E[X] = −1
4+0+1
4
Step 4: Combining the terms, we find:
E[X] = 0
Therefore, the expected value of the random variable Xis 0.
10
Question 14
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
2, P (X=−2) = 1
6, P (X= 3) = 1
3
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
by E(X), is given by:
E(X) = X
x
x·P(X=x)
Step 2: Apply this formula to find the expected value of Xin this case:
E(X) = 1 ·1
2+ (−2) ·1
6+ 3 ·1
3
Step 3: Simplify the expression:
E(X) = 1
2−1
3+ 1
E(X) = 3
6−2
6+ 2
E(X) = 1
6+4
6
Step 4: Finalize the computation:
E(X) = 5
6
Therefore, the expected value of the random variable Xis 5
6.
Question 15
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−2) = 1
8, P (X= 0) = 5
16, P (X= 3) = 3
8, P (X=k) = 1
k2,for k= 4,5,6, . . .
Determine the expected value of X.
11
Solution
Step 1: First, we recall that the expected value of a discrete random variable X
is given by:
E(X) = X
k
k·P(X=k)
Step 2: We evaluate the expected value by summing over all possible values
of X:
E(X)=(−2) 1
8+ (0) 5
16+ (3) 3
8+
∞
X
k=4
k·1
k2
Step 3: Simplifying the expression, we have:
E(X) = −2
8+0+9
8+
∞
X
k=4
1
k
Step 4: The summation can be simplified as a harmonic series, which di-
verges. Therefore, the expected value E(X) is infinite.
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
x1 2
3
P(X=x) 0.3 0.5
0.2
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: We substitute the values from the probability distribution into the
formula:
E(X)=1·0.3+2·0.5+3·0.2
Step 3: Calculate the expected value:
E(X) = 0.3+1.0+0.6
12
E(X)=1.9
Therefore, the expected value of Xis 1.9.
Question 17
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
6, P (X= 0) = 2
3, P (X= 1) = 1
6
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 given probabilities into the formula for the expected
value:
E(X)=(−1) ·1
6+ (0) ·2
3+ (1) ·1
6
Step 3: Simplify the expression:
E(X) = −1
6+0+1
6= 0
Step 4: Therefore, the expected value of the random variable Xis 0 .
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) = 0.2, P (X= 2) = 0.2
Calculate the expected value of X.
13
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: Substitute the values of Xand P(X) into the formula to find the
expected value:
E[X] = (−2) ·0.1 + (0) ·0.5 + (1) ·0.2 + (2) ·0.2
Step 3: Calculate the expected value:
E[X] = −0.2+0+0.2+0.4
E[X]=0.4
Step 4: Therefore, the expected value of the random variable Xis 0.4.
Question 19
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=k) = 3
21
2k+1
Find the expected value of X.
Solution
Step 1: First, let’s recall the formula for the expected value of a discrete random
variable X:
E(X) = X
k
k·P(X=k)
Step 2: Using the given probability mass function, we have:
E(X) =
∞
X
k=0
k·3
21
2k+1
Step 3: Simplifying the expression, we get:
E(X) = 3
2
∞
X
k=0
k·1
2k+1
14
Step 4: We can rewrite k·1
2k+1 as the derivative with respect to pof
1
2k+1:
d
dp 1
2k+1
=d
dp2−k−1=−(k+ 1)2−k−2
Step 5: So, the expected value can be written as:
E(X) = 3
2
d
dp
∞
X
k=0
2−k−1
Step 6: Simplifying further, we find:
E(X) = 3
2
d
dp 1/2
1−1/2
Step 7: Finally, we obtain the expected value of Xas:
E(X) = 3
2
d
dp(1) = 3
2·0=0
Therefore, the expected value of the discrete random variable Xis 0.
Question 20
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.2, P (X= 2) = 0.3, P (X= 4) = 0.4
Calculate the expected value of X.
Solution
Step 1: First, we recall the definition of the expected value of a discrete random
variable X:
E(X) = X
x
x·P(X=x)
Step 2: Given the probability mass function of X, we can calculate the
expected value as follows:
E(X) = (−2) ·0.1+0·0.2+2·0.3+4·0.4
Step 3: Simplifying the expression, we get:
15
E(X) = −0.2+0+0.6+1.6
Step 4: Combining the terms, we find:
E(X) = 2
Therefore, the expected value of the random variable Xis 2.
Question 21
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X=−1) = 0.2, P (X= 0) = 0.3, P (X= 1) = 0.2, 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 cal-
culated as:
E(X) = X
x
x·P(X=x).
Step 2: Substitute the given probabilities into the formula:
E(X) = (−2) ·0.1+(−1) ·0.2 + (0) ·0.3 + (1) ·0.2 + (2) ·0.2.
Step 3: Simplify the expression:
E(X) = −0.2−0.2+0+0.2+0.4.
Step 4: Calculate the sum:
E(X) = 0.2.
Therefore, the expected value of the random variable Xis 0.
Question 22
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−2) = 0.25, P (X= 1) = 0.1,and P(X= 3) = 0.65
Determine the expected value of X.
16
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)
where the sum is taken over all possible values of X.
Step 2: Substitute the given probabilities for X=−2,1,3 into the formula:
E(X)=(−2) ·0.25 + (1) ·0.1 + (3) ·0.65
Step 3: Calculate the expected value:
E(X) = −0.5+0.1+1.95 = 1.55
Therefore, the expected value of Xis 1.55.
Question 23
Question
Let Xbe a discrete random variable with probability mass function given by
P(X=−2) = 0.2, P (X= 0) = 0.5, P (X= 2) = 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
E(X) = Xxi·P(X=xi),
where the sum is taken over all possible values xithat Xcan take.
Step 2: In this case, the possible values that Xcan take are -2, 0, and 2.
So, we have
E(X) = (−2) ·0.2 + (0) ·0.5 + (2) ·0.3.
Step 3: Calculate the expected value E(X):
E(X)=(−2)(0.2) + (0)(0.5) + (2)(0.3) = −0.4+0+0.6.
E(X)=0.2.
Step 4: Therefore, the expected value of the random variable Xis E(X) =
0.2.
17
Question 24
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
2, P (X= 2) = 1
3, P (X= 3) = 1
6
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: Given that Xis a discrete random variable with probability mass
function:
P(X= 1) = 1
2, P (X= 2) = 1
3, P (X= 3) = 1
6
Step 3: We can calculate the expected value of Xby substituting the values
of xand P(X=x) into the formula for expected value:
E(X)=1·1
2+ 2 ·1
3+ 3 ·1
6
Step 4: Simplifying the expression:
E(X) = 1
2+2
3+3
6=3
6+4
6+3
6=10
6=5
3
Therefore, the expected value of the random variable Xis 5
3.
Question 25
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−1) = 1
3, P (X= 3) = 1
6, P (X= 5) = 1
2
Find the expected value E(X) of the random variable X.
18
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 their corresponding probabilities into
the formula for expected value:
E(X)=(−1) 1
3+ (3) 1
6+ (5) 1
2
Step 3: Calculate the expected value:
E(X) = −1
3+1
2+5
2
E(X) = −2
6+3
6+15
6
E(X) = 16
6
Step 4: Simplify the fraction to get the final answer:
E(X) = 8
3
Question 26
Question
Let Xbe a discrete random variable with the following probability distribution:
X012
P(X) 0.4 0.3 0.3
Calculate the expected value of X,E(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: Calculate E(X) by multiplying each value of Xby its corresponding
probability and adding them together:
E(X)=0·0.4+1·0.3+2·0.3
E(X) = 0 + 0.3+0.6
E(X)=0.9
Step 3: Therefore, the expected value of Xis 0.9.
19
Question 27
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.2, P (X= 0) = 0.3, P (X= 1) = 0.5
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: Plug in the values from the given probability mass function:
E(X)=(−1) ·0.2 + (0) ·0.3 + (1) ·0.5
Step 3: Simplify the expression:
E(X) = −0.2+0+0.5
Step 4: Calculate the final expected value:
E(X)=0.3
Therefore, the expected value of the random variable Xis 0.3.
Question 28
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = 1
2k
Find the expected value of X.
20
Solution
Step 1: To find the expected value of X, we use the formula:
E(X) = X
k
k·P(X=k)
Step 2: First, let’s find P(X=k) for all possible values of k.
For k= 1:
P(X= 1) = 1
21=1
2
For k= 2:
P(X= 2) = 1
22=1
4
For k= 3:
P(X= 3) = 1
23=1
8
Similarly, for k= 4:
P(X= 4) = 1
24=1
16
And so on.
Step 3: Now, substitute the probabilities back into the formula for expected
value:
E(X)=1·1
2+ 2 ·1
4+ 3 ·1
8+ 4 ·1
16 +. . .
Step 4: Simplify the expression:
E(X) = 1
2+2
4+3
8+4
16 +. . .
E(X) = 1
2+1
2+3
8+1
4+. . .
Step 5: Continuing this pattern, we find:
E(X) =
∞
X
k=1
k
2k
Step 6: The above series is an arithmetic-geometric series and can be sim-
plified using the formula:
S=
∞
X
n=1
nrn−1=r
(1 −r)2
Step 7: Applying the formula, we get:
E(X) =
1
2
(1 −1
2)2=1
(1/2)2= 2
Therefore, the expected value of Xis 2.
21
Question 29
Question
Let Xbe a discrete random variable with the following probability distribution:
P(X=−1) = 0.2, P (X= 0) = 0.3, P (X= 1) = 0.4, P (X= 2) = 0.1.
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 the sum is taken over all possible values of X.
Step 2: Substitute the given probabilities into the formula:
E(X)=(−1) ·0.2 + (0) ·0.3 + (1) ·0.4 + (2) ·0.1.
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.4+0.2 = 0.4.
Therefore, the expected value of the random variable Xis 0.4.
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 3
P(X)1
4
1
2
1
4
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
i
xi·P(X=xi)
where xiare the possible values that Xcan take on, and P(X=xi) are the
corresponding probabilities.
Step 2: In this case, we have:
22
E(X)=(−2) ·1
4+ (0) ·1
2+ (3) ·1
4
Step 3: Simplify the expression:
E(X) = −1
2+0+3
4
E(X) = 1
4
Step 4: Therefore, the expected value of the random variable Xis 1
4.
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
18, P (X= 0) = 1
3, P (X= 3) = 1
6, P (X= 5) = 1
9, P (X= 7) = 1
6.
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),
where the sum is taken over all possible values of X.
Step 2: Substitute the values of Xand their probabilities into the formula
for the expected value:
E[X] = (−2) ·1
18 + (0) ·1
3+ (3) ·1
6+ (5) ·1
9+ (7) ·1
6.
Step 3: Calculate the expected value:
E[X] = −2
18 +0+3
6+5
9+7
6=−1
2.
Therefore, the expected value of Xis −1
2.
23
Question 32
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.3, P (X= 2) = 0.2, P (X= 3) = 0.1, P (X= 4) = 0.4
Find the expected value E(X) of the random variable X.
Solution
Step 1: Recall that the expected value E(X) of a discrete random variable X
is given by:
E(X) = X
all x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) into the formula and
calculate the expected value E(X).
E(X) = 1 ×0.3+2×0.2+3×0.1+4×0.4
Step 3: Perform the calculations to find the expected value E(X).
E(X) = 0.3+0.4+0.3+1.6=2.6
Therefore, the expected value of the random variable Xis E(X) = 2.6.
Question 33
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
6, P (X= 2) = 1
3, P (X= 3) = 1
4, P (X= 4) = 1
12
Calculate the expected value of X.
24
Solution
Step 1: 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:
E(X)=1·1
6+ 2 ·1
3+ 3 ·1
4+ 4 ·1
12
Step 3: Simplify the expression:
E(X) = 1
6+2
3+3
4+1
3
Step 4: Find a common denominator and add the fractions:
E(X) = 2
12 +8
12 +9
12 +3
12
Step 5: Combine the fractions:
E(X) = 22
12
Step 6: Simplify the fraction:
E(X) = 11
6
Therefore, the expected value of the random variable Xis 11
6.
Question 34
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−2) = 1
6, P (X= 0) = 1
3, P (X= 2) = 1
2
Find 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 xthat Xcan take.
25
Step 2: Substitute the values of xand P(X=x) into the formula to find
the expected value:
E(X)=(−2) ·P(X=−2) + (0) ·P(X= 0) + (2) ·P(X= 2)
E(X)=(−2) ·1
6+ (0) ·1
3+ (2) ·1
2
Step 3: Simplify the expression:
E(X) = −2
6+0+2
2
E(X) = −1
3+ 1
E(X) = 2
3
Therefore, the expected value of the random variable Xis 2
3.
Question 35
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) =
3
8for k=−2
1
8for k=−1
1
4for k= 0
1
8for k= 1
1
8for k= 2
0 otherwise
Calculate E(X), the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis defined as:
E(X) = X
k
k·P(X=k)
Step 2: Substitute the values from the probability mass function into the
formula:
E(X)=(−2) ·3
8+ (−1) ·1
8+ 0 ·1
4+ 1 ·1
8+ 2 ·1
8
Step 3: Simplify the expression:
26
Question 3
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=x) =
0.2 if x= 0
0.3 if x= 1
0.1 if x= 2
0.4 if x= 3
Determine the expected value of X.
Solution
Step 1: The expected value E(X) of a discrete random variable Xis defined as:
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)=0·0.2+1·0.3+2·0.1+3·0.4
Step 3: Simplifying the expression:
E(X) = 0 + 0.3+0.2+1.2
Step 4: Thus, the expected value of Xis:
E(X) = 1.7
Therefore, the expected value of Xis 1.7.
Question 4
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 4
P(X)1
6
1
2
1
3
Calculate the expected value of X.
3
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of X.
Step 2: Using the given probability distribution of X, we can calculate the
expected value as:
E(X)=(−2) ·1
6+ (0) ·1
2+ (4) ·1
3
Step 3: Simplifying the expression, we get:
E(X) = −2
6+0+4
3
E(X) = −1
3+4
3
E(X) = 3
3= 1
Therefore, the expected value of the random variable Xis 1.
Question 5
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
6, P (X= 0) = 1
3, P (X= 2) = 1
2.
Calculate the expected value of X.
Solution
Step 1: First, 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 given probability mass function into
the formula:
E(X)=(−2) ·1
6+ (0) ·1
3+ (2) ·1
2.
Step 3: Simplify the expression:
E(X) = −1
3+ 0 + 1 = 2
3.
Step 4: Therefore, the expected value of the random variable Xis 2
3.
4
Question 6
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2
P(X=x) 0.3 0.5 0.2
Find the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis defined as:
E(X) = X
all x
x·P(X=x)
Step 2: Using the given probability distribution, we can calculate the ex-
pected value of Xas:
E(X)=0·0.3+1·0.5+2·0.2
Step 3: Simplifying the expression, we get:
E(X) = 0 + 0.5+0.4
E(X)=0.9
Therefore, the expected value of the random variable Xis 0.9.
Question 7
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
8, P (X= 0) = 3
8, P (X= 2) = 1
4, P (X= 4) = 1
4.
Calculate the expected value of X.
5
Solution
Step 1: To find the expected value of a discrete random variable X, we use the
formula:
E(X) = X
all x
x·P(X=x).
Step 2: Given the probability mass function of X, we can substitute in the
values of xand P(X=x):
E(X)=(−2) ·1
8+ (0) ·3
8+ (2) ·1
4+ (4) ·1
4.
Step 3: Simplifying the expression:
E(X) = −2
8+0+2
4+4
4=−1
4+1
2+ 1.
Step 4: Further simplifying:
E(X) = 1
4+2
4+4
4=7
4.
Therefore, the expected value of the random variable Xis 7
4.
Question 8
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=x) = (1
2x+1 ,if x= 0,1,2, . . .
0,otherwise
Calculate the expected value of X.
Solution
Step 1: To find the expected value of a discrete random variable X, denoted by
E[X], we use the formula:
E[X] = X
x
x·P(X=x)
Step 2: Given the probability mass function of X, we need to calculate E[X]
as follows:
E[X] =
∞
X
x=0
x·1
2x+1
6
Step 3: Let’s simplify the expression before calculating the sum. Notice that
x·1
2x+1 can be rewritten as x
2x. So we have:
E[X] =
∞
X
x=0
x
2x
Step 4: To calculate this sum, we can differentiate a known power series.
Let’s consider the power series expansion of the function f(t) = P∞
x=0 tx. This
is a geometric series that converges to 1
1−tfor |t|<1.
Step 5: Differentiating both sides with respect to t, we get:
f′(t) =
∞
X
x=0
x·tx−1=d
dt 1
1−t
Step 6: Simplifying the right side and using the chain rule, we have:
∞
X
x=0
x·tx−1=1
(1 −t)2
Step 7: Setting t=1
2, we find:
∞
X
x=0
x·1
2x−1
=1
1−1
22= 4
Step 8: Therefore, the expected value of the random variable Xis E[X] = 4.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−3) = 0.1, P (X= 0) = 0.3, P (X= 2) = 0.4, P (X= 5) = 0.2
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)
where the sum is taken over all possible values of X.
7
Step 2: In this case, the possible values of Xare -3, 0, 2, and 5.
Step 3: We can now calculate the expected value:
E(X)=(−3)(0.1) + (0)(0.3) + (2)(0.4) + (5)(0.2)
Step 4: Simplifying, we get:
E(X) = −0.3+0+0.8 + 1 = 1.5
Step 5: Therefore, the expected value of the random variable Xis 1.5.
Question 10
Question
Let Xbe a discrete random variable with the following probability distribution:
x P (X=x)
1 0.2
2 0.3
3 0.1
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 the formula:
E(X) = Xx·P(X=x)
Step 2: We will substitute the values from the probability distribution of X
into the formula to calculate the expected value:
E(X)=1·0.2+2·0.3+3·0.1+5·0.4
Step 3: Calculate the expected value:
E(X) = 0.2+0.6+0.3 + 2 = 3.1
Therefore, the expected value of the random variable Xis 3.1.
Question 11
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
8
P(X=−2) = 1
9, P (X= 0) = 1
3, P (X= 1) = 1
6, P (X= 3) = 1
4
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
i
xi·P(X=xi)
where the sum is taken over all possible values xithat Xcan take.
Step 2: Substitute the values of Xand their corresponding probabilities into
the formula:
E(X) = (−2) 1
9+ (0) 1
3+ (1) 1
6+ (3) 1
4
Step 3: Simplify the expression by multiplying and adding the terms:
E(X) = −2
9+0+1
6+3
4
E(X) = −8
36 +0+ 6
36 +27
36
E(X) = 25
36
Step 4: Therefore, the expected value of the random variable Xis 25
36 .
Question 12
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
4, P (X= 0) = 1
2, P (X= 1) = 1
4
Calculate the expected value of X.
9
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: Substitute the given probabilities into the formula.
E(X)=(−1) ·1
4+ (0) ·1
2+ (1) ·1
4
Step 3: Simplify the expression.
E(X) = −1
4+0+1
4= 0
Step 4: Therefore, the expected value of the random variable Xis 0 .
Question 13
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−1) = 1
4, P (X= 0) = 1
2, P (X= 1) = 1
4
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis defined
as:
E[X] = X
all x
x·P(X=x)
Step 2: Using the given probability mass function for X, we have:
E[X]=(−1) ·1
4+ (0) ·1
2+ (1) ·1
4
Step 3: Simplifying the expression, we get:
E[X] = −1
4+0+1
4
Step 4: Combining the terms, we find:
E[X] = 0
Therefore, the expected value of the random variable Xis 0.
10
Question 14
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
2, P (X=−2) = 1
6, P (X= 3) = 1
3
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
by E(X), is given by:
E(X) = X
x
x·P(X=x)
Step 2: Apply this formula to find the expected value of Xin this case:
E(X) = 1 ·1
2+ (−2) ·1
6+ 3 ·1
3
Step 3: Simplify the expression:
E(X) = 1
2−1
3+ 1
E(X) = 3
6−2
6+ 2
E(X) = 1
6+4
6
Step 4: Finalize the computation:
E(X) = 5
6
Therefore, the expected value of the random variable Xis 5
6.
Question 15
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−2) = 1
8, P (X= 0) = 5
16, P (X= 3) = 3
8, P (X=k) = 1
k2,for k= 4,5,6, . . .
Determine the expected value of X.
11
Solution
Step 1: First, we recall that the expected value of a discrete random variable X
is given by:
E(X) = X
k
k·P(X=k)
Step 2: We evaluate the expected value by summing over all possible values
of X:
E(X)=(−2) 1
8+ (0) 5
16+ (3) 3
8+
∞
X
k=4
k·1
k2
Step 3: Simplifying the expression, we have:
E(X) = −2
8+0+9
8+
∞
X
k=4
1
k
Step 4: The summation can be simplified as a harmonic series, which di-
verges. Therefore, the expected value E(X) is infinite.
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
x1 2
3
P(X=x) 0.3 0.5
0.2
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: We substitute the values from the probability distribution into the
formula:
E(X)=1·0.3+2·0.5+3·0.2
Step 3: Calculate the expected value:
E(X) = 0.3+1.0+0.6
12
E(X)=1.9
Therefore, the expected value of Xis 1.9.
Question 17
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
6, P (X= 0) = 2
3, P (X= 1) = 1
6
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 given probabilities into the formula for the expected
value:
E(X)=(−1) ·1
6+ (0) ·2
3+ (1) ·1
6
Step 3: Simplify the expression:
E(X) = −1
6+0+1
6= 0
Step 4: Therefore, the expected value of the random variable Xis 0 .
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) = 0.2, P (X= 2) = 0.2
Calculate the expected value of X.
13
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: Substitute the values of Xand P(X) into the formula to find the
expected value:
E[X] = (−2) ·0.1 + (0) ·0.5 + (1) ·0.2 + (2) ·0.2
Step 3: Calculate the expected value:
E[X] = −0.2+0+0.2+0.4
E[X]=0.4
Step 4: Therefore, the expected value of the random variable Xis 0.4.
Question 19
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=k) = 3
21
2k+1
Find the expected value of X.
Solution
Step 1: First, let’s recall the formula for the expected value of a discrete random
variable X:
E(X) = X
k
k·P(X=k)
Step 2: Using the given probability mass function, we have:
E(X) =
∞
X
k=0
k·3
21
2k+1
Step 3: Simplifying the expression, we get:
E(X) = 3
2
∞
X
k=0
k·1
2k+1
14
Step 4: We can rewrite k·1
2k+1 as the derivative with respect to pof
1
2k+1:
d
dp 1
2k+1
=d
dp2−k−1=−(k+ 1)2−k−2
Step 5: So, the expected value can be written as:
E(X) = 3
2
d
dp
∞
X
k=0
2−k−1
Step 6: Simplifying further, we find:
E(X) = 3
2
d
dp 1/2
1−1/2
Step 7: Finally, we obtain the expected value of Xas:
E(X) = 3
2
d
dp(1) = 3
2·0=0
Therefore, the expected value of the discrete random variable Xis 0.
Question 20
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.2, P (X= 2) = 0.3, P (X= 4) = 0.4
Calculate the expected value of X.
Solution
Step 1: First, we recall the definition of the expected value of a discrete random
variable X:
E(X) = X
x
x·P(X=x)
Step 2: Given the probability mass function of X, we can calculate the
expected value as follows:
E(X) = (−2) ·0.1+0·0.2+2·0.3+4·0.4
Step 3: Simplifying the expression, we get:
15
E(X) = −0.2+0+0.6+1.6
Step 4: Combining the terms, we find:
E(X) = 2
Therefore, the expected value of the random variable Xis 2.
Question 21
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X=−1) = 0.2, P (X= 0) = 0.3, P (X= 1) = 0.2, 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 cal-
culated as:
E(X) = X
x
x·P(X=x).
Step 2: Substitute the given probabilities into the formula:
E(X) = (−2) ·0.1+(−1) ·0.2 + (0) ·0.3 + (1) ·0.2 + (2) ·0.2.
Step 3: Simplify the expression:
E(X) = −0.2−0.2+0+0.2+0.4.
Step 4: Calculate the sum:
E(X) = 0.2.
Therefore, the expected value of the random variable Xis 0.
Question 22
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−2) = 0.25, P (X= 1) = 0.1,and P(X= 3) = 0.65
Determine the expected value of X.
16
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)
where the sum is taken over all possible values of X.
Step 2: Substitute the given probabilities for X=−2,1,3 into the formula:
E(X)=(−2) ·0.25 + (1) ·0.1 + (3) ·0.65
Step 3: Calculate the expected value:
E(X) = −0.5+0.1+1.95 = 1.55
Therefore, the expected value of Xis 1.55.
Question 23
Question
Let Xbe a discrete random variable with probability mass function given by
P(X=−2) = 0.2, P (X= 0) = 0.5, P (X= 2) = 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
E(X) = Xxi·P(X=xi),
where the sum is taken over all possible values xithat Xcan take.
Step 2: In this case, the possible values that Xcan take are -2, 0, and 2.
So, we have
E(X) = (−2) ·0.2 + (0) ·0.5 + (2) ·0.3.
Step 3: Calculate the expected value E(X):
E(X)=(−2)(0.2) + (0)(0.5) + (2)(0.3) = −0.4+0+0.6.
E(X)=0.2.
Step 4: Therefore, the expected value of the random variable Xis E(X) =
0.2.
17
Question 24
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
2, P (X= 2) = 1
3, P (X= 3) = 1
6
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: Given that Xis a discrete random variable with probability mass
function:
P(X= 1) = 1
2, P (X= 2) = 1
3, P (X= 3) = 1
6
Step 3: We can calculate the expected value of Xby substituting the values
of xand P(X=x) into the formula for expected value:
E(X)=1·1
2+ 2 ·1
3+ 3 ·1
6
Step 4: Simplifying the expression:
E(X) = 1
2+2
3+3
6=3
6+4
6+3
6=10
6=5
3
Therefore, the expected value of the random variable Xis 5
3.
Question 25
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−1) = 1
3, P (X= 3) = 1
6, P (X= 5) = 1
2
Find the expected value E(X) of the random variable X.
18
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 their corresponding probabilities into
the formula for expected value:
E(X)=(−1) 1
3+ (3) 1
6+ (5) 1
2
Step 3: Calculate the expected value:
E(X) = −1
3+1
2+5
2
E(X) = −2
6+3
6+15
6
E(X) = 16
6
Step 4: Simplify the fraction to get the final answer:
E(X) = 8
3
Question 26
Question
Let Xbe a discrete random variable with the following probability distribution:
X012
P(X) 0.4 0.3 0.3
Calculate the expected value of X,E(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: Calculate E(X) by multiplying each value of Xby its corresponding
probability and adding them together:
E(X)=0·0.4+1·0.3+2·0.3
E(X) = 0 + 0.3+0.6
E(X)=0.9
Step 3: Therefore, the expected value of Xis 0.9.
19
Question 27
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.2, P (X= 0) = 0.3, P (X= 1) = 0.5
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: Plug in the values from the given probability mass function:
E(X)=(−1) ·0.2 + (0) ·0.3 + (1) ·0.5
Step 3: Simplify the expression:
E(X) = −0.2+0+0.5
Step 4: Calculate the final expected value:
E(X)=0.3
Therefore, the expected value of the random variable Xis 0.3.
Question 28
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = 1
2k
Find the expected value of X.
20
Solution
Step 1: To find the expected value of X, we use the formula:
E(X) = X
k
k·P(X=k)
Step 2: First, let’s find P(X=k) for all possible values of k.
For k= 1:
P(X= 1) = 1
21=1
2
For k= 2:
P(X= 2) = 1
22=1
4
For k= 3:
P(X= 3) = 1
23=1
8
Similarly, for k= 4:
P(X= 4) = 1
24=1
16
And so on.
Step 3: Now, substitute the probabilities back into the formula for expected
value:
E(X)=1·1
2+ 2 ·1
4+ 3 ·1
8+ 4 ·1
16 +. . .
Step 4: Simplify the expression:
E(X) = 1
2+2
4+3
8+4
16 +. . .
E(X) = 1
2+1
2+3
8+1
4+. . .
Step 5: Continuing this pattern, we find:
E(X) =
∞
X
k=1
k
2k
Step 6: The above series is an arithmetic-geometric series and can be sim-
plified using the formula:
S=
∞
X
n=1
nrn−1=r
(1 −r)2
Step 7: Applying the formula, we get:
E(X) =
1
2
(1 −1
2)2=1
(1/2)2= 2
Therefore, the expected value of Xis 2.
21
Question 29
Question
Let Xbe a discrete random variable with the following probability distribution:
P(X=−1) = 0.2, P (X= 0) = 0.3, P (X= 1) = 0.4, P (X= 2) = 0.1.
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 the sum is taken over all possible values of X.
Step 2: Substitute the given probabilities into the formula:
E(X)=(−1) ·0.2 + (0) ·0.3 + (1) ·0.4 + (2) ·0.1.
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.4+0.2 = 0.4.
Therefore, the expected value of the random variable Xis 0.4.
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 3
P(X)1
4
1
2
1
4
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
i
xi·P(X=xi)
where xiare the possible values that Xcan take on, and P(X=xi) are the
corresponding probabilities.
Step 2: In this case, we have:
22
E(X)=(−2) ·1
4+ (0) ·1
2+ (3) ·1
4
Step 3: Simplify the expression:
E(X) = −1
2+0+3
4
E(X) = 1
4
Step 4: Therefore, the expected value of the random variable Xis 1
4.
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
18, P (X= 0) = 1
3, P (X= 3) = 1
6, P (X= 5) = 1
9, P (X= 7) = 1
6.
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),
where the sum is taken over all possible values of X.
Step 2: Substitute the values of Xand their probabilities into the formula
for the expected value:
E[X] = (−2) ·1
18 + (0) ·1
3+ (3) ·1
6+ (5) ·1
9+ (7) ·1
6.
Step 3: Calculate the expected value:
E[X] = −2
18 +0+3
6+5
9+7
6=−1
2.
Therefore, the expected value of Xis −1
2.
23
Question 32
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.3, P (X= 2) = 0.2, P (X= 3) = 0.1, P (X= 4) = 0.4
Find the expected value E(X) of the random variable X.
Solution
Step 1: Recall that the expected value E(X) of a discrete random variable X
is given by:
E(X) = X
all x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) into the formula and
calculate the expected value E(X).
E(X) = 1 ×0.3+2×0.2+3×0.1+4×0.4
Step 3: Perform the calculations to find the expected value E(X).
E(X) = 0.3+0.4+0.3+1.6=2.6
Therefore, the expected value of the random variable Xis E(X) = 2.6.
Question 33
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
6, P (X= 2) = 1
3, P (X= 3) = 1
4, P (X= 4) = 1
12
Calculate the expected value of X.
24
Solution
Step 1: 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:
E(X)=1·1
6+ 2 ·1
3+ 3 ·1
4+ 4 ·1
12
Step 3: Simplify the expression:
E(X) = 1
6+2
3+3
4+1
3
Step 4: Find a common denominator and add the fractions:
E(X) = 2
12 +8
12 +9
12 +3
12
Step 5: Combine the fractions:
E(X) = 22
12
Step 6: Simplify the fraction:
E(X) = 11
6
Therefore, the expected value of the random variable Xis 11
6.
Question 34
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−2) = 1
6, P (X= 0) = 1
3, P (X= 2) = 1
2
Find 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 xthat Xcan take.
25
Step 2: Substitute the values of xand P(X=x) into the formula to find
the expected value:
E(X)=(−2) ·P(X=−2) + (0) ·P(X= 0) + (2) ·P(X= 2)
E(X)=(−2) ·1
6+ (0) ·1
3+ (2) ·1
2
Step 3: Simplify the expression:
E(X) = −2
6+0+2
2
E(X) = −1
3+ 1
E(X) = 2
3
Therefore, the expected value of the random variable Xis 2
3.
Question 35
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) =
3
8for k=−2
1
8for k=−1
1
4for k= 0
1
8for k= 1
1
8for k= 2
0 otherwise
Calculate E(X), the expected value of X.
Solution
Step 1: The expected value of a discrete random variable Xis defined as:
E(X) = X
k
k·P(X=k)
Step 2: Substitute the values from the probability mass function into the
formula:
E(X)=(−2) ·3
8+ (−1) ·1
8+ 0 ·1
4+ 1 ·1
8+ 2 ·1
8
Step 3: Simplify the expression:
26
E(X) = −6
8−1
8+0+1
8+2
8=−4
8=−1
2
Therefore, the expected value of the random variable Xis −1
2.
27
