MATH 350 - DISCRETE
MATHEMATICS - Discrete random
variables and expected value
Question Bank - Set 2
Liberty University
Question 1
Question
Let Xbe a discrete random variable with the following probability distribution:
X0 1 2 3
P(X) 0.1 0.2 0.3 0.4
Find the expected value of X.
Solution
Step 1: To find the expected value of X, denoted by E(X), we use the formula:
E(X) = X
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) are the probabilities
associated with those values.
Step 2: Substitute the values of xiand P(X=xi) into the formula:
E(X)=0·0.1+1·0.2+2·0.3+3·0.4
Step 3: Calculate the expected value:
E(X) = 0 + 0.2+0.6+1.2 = 2
Therefore, the expected value of the random variable Xis 2.
Question 2
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.2, P (X= 2) = 0.3, P (X= 3) = 0.1, P (X= 4) = 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
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: Calculate the expected value E(X) using the given probability mass
function:
E(X) = 1 ·0.2+2·0.3+3·0.1+4·0.4.
Step 3: Compute the expected value of X:
E(X)=0.2+0.6+0.3+1.6=2.7.
Therefore, the expected value of the random variable Xis 2.7.
Question 3
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
0.2 if x= 1,
0.3 if x= 2,
0.1 if x= 3,
0.2 if x= 4,
0.2 if x= 5.
Calculate the expected value of X.
2
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=x) into the formula:
E(X)=1·0.2+2·0.3+3·0.1+4·0.2+5·0.2
Step 3: Perform the calculations:
E(X)=0.2+0.6+0.3+0.8 + 1 = 3
Step 4: Therefore, the expected value of the random variable Xis 3.
Question 4
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= 1) = 0.2, P (X= 2) = 0.4
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 values from the provided probability mass function
into the formula:
E(X)=(−2) ·0.1 + (0) ·0.3 + (1) ·0.2 + (2) ·0.4
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.2+0.8 = 0.8
Therefore, the expected value of the random variable Xis 0.8.
3
Question 5
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 3) = 1
2, P (X= 5) = 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) = Xxi·P(X=xi)
where the sum is taken over all possible values xiof X.
Step 2: In this case, we have X={1,3,5}and their corresponding proba-
bilities:
P(X= 1) = 1
4, P (X= 3) = 1
2, P (X= 5) = 1
4
Step 3: Now, we can calculate the expected value using the formula:
E(X) = 1 ·1
4+ 3 ·1
2+ 5 ·1
4
Step 4: Simplifying the expression, we get:
E(X) = 1
4+3
2+5
4
E(X) = 1
4+6
4+5
4
E(X) = 12
4
Step 5: Finally, we have:
E(X) = 12
4= 3
Therefore, the expected value of the random variable Xis 3.
Question 6
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.2, P (X= 2) = 0.3, P (X= 3) = 0.4, P (X= 4) = 0.1.
Find the expected value of X.
4
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 xthat Xcan take on.
Step 2: Substitute the given probabilities into the formula for expected value:
E(X) = 1 ·0.2+2·0.3+3·0.4+4·0.1.
Step 3: Calculate the expected value:
E(X)=0.2+0.6+1.2+0.4=2.4.
Therefore, the expected value of the random variable Xis 2.4 .
Question 7
Question
Let Xbe a discrete random variable with the following probability distribution:
X123
P(X) 0.3 0.5 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: We can now substitute the values from the probability distribution
into the formula for expected value:
E(X)=1·0.3+2·0.5+3·0.2
Step 3: Calculate the products and sum them up:
E(X) = 0.3+1.0+0.6
Step 4: Finally, sum up the values to get the expected value of X:
E(X)=1.9
Therefore, the expected value of the random variable Xis 1.9.
5
Question 8
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.2, P (X= 0) = 0.5, P (X= 1) = 0.3.
Calculate the expected value of X.
Solution
Step 1: Find the expected value of Xusing 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:
E(X)=(−1)(0.2) + (0)(0.5) + (1)(0.3).
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.3 = 0.1.
Therefore, the expected value of random variable Xis 0.1.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.25, P (X= 1) = 0.1, P (X= 3) = 0.2, P (X= 4) = 0.45.
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 of X.
6
Step 2: Calculate the expected value E(X) using the given probability mass
function:
E(X)=(−2) ·0.25 + (1) ·0.1 + (3) ·0.2 + (4) ·0.45
Step 3: Multiply each value by its probability and sum them up:
E(X) = −0.5+0.1+0.6+1.8
Step 4: Calculate the final expected value:
E(X)=2.0
Question 10
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 3
P(X) 0.2 0.5 0.3
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 of Xand P(X=xi) are the corresponding
probabilities.
Step 2: Substitute the values from the probability distribution into the for-
mula:
E(X)=(−2) ·0.2 + (0) ·0.5 + (3) ·0.3
Step 3: Perform the calculations:
E(X) = −0.4+0+0.9
E(X)=0.5
Therefore, the expected value of the discrete random variable Xis 0.5.
7
Question 11
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 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 from the probability mass function into the
formula:
E(X)=1·1
3+ 2 ·1
6+ 3 ·1
2
Step 3: Simplify the expression:
E(X) = 1
3+1
3+3
2
Step 4: Combine the fractions:
E(X) = 2
3+3
2=4
6+9
6=13
6
Step 5: Therefore, the expected value of the random variable Xis 13
6.
Question 12
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= 1) = 0.2, P (X= 3) = 0.4.
Find the expected value of X.
8
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: Calculate the expected value by multiplying each value of Xby its
corresponding probability and summing them up:
E(X)=(−2)(0.1) + (0)(0.3) + (1)(0.2) + (3)(0.4)
E(X) = −0.2+0+0.2+1.2
E(X) = 1.2
Therefore, the expected value of the random variable Xis 1.2.
Question 13
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2 3 4
P(X)1
4
1
4
1
4
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
x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) from the probability
distribution into the formula:
E(X)=1·1
4+ 2 ·1
4+ 3 ·1
4+ 4 ·1
4
Step 3: Simplify the expression:
E(X) = 1
4+2
4+3
4+4
4
E(X) = 10
4
9
E(X)=2.5
Step 4: Therefore, the expected value of the random variable Xis 2.5.
Question 14
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2 3
P(X) 0.2 0.3p
If E[X] = 2.2, find the value of p.
Solution
Step 1: The expected value E[X] 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 xthat Xcan take.
Given that E[X]=2.2 and the probability distribution of X, we can write:
E[X]=1·0.2+2·0.3+3·p
Step 2: Simplify the equation:
2.2=0.2+0.6+3p
2.2=0.8+3p
1.4=3p
p=1.4
3= 0.4667
Therefore, the value of pis 0.4667 .
10
Question 15
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,and P(X=k) = 0.4 for some k∈Z.
Find the value of kand calculate the expected value of X.
Solution
Step 1: Find the value of k Since the sum of all probabilities must equal 1,
we have:
0.1+0.2+0.3+0.4 = 1 =⇒1.0 = 1
Therefore, the missing probability is:
0.4 = P(X=k)
0.4=1−0.1−0.2−0.3
0.4=0.4
Hence, kmust be equal to 1.
Step 2: Calculate the expected value of X The expected value of a
discrete random variable Xis given by:
E(X) = Xx·P(X=x)
Substitute the values from the probability mass function:
E(X)=(−2)(0.1) + (0)(0.2) + (3)(0.3) + (1)(0.4)
E(X) = −0.2+0+0.9+0.4
E(X)=1.1
Therefore, the expected value of Xis 1.
11
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 1 3
P(X)k1
3
1
2
Find the value of kand calculate the expected value of X.
Solution
Step 1: Find the value of kusing the fact that the probabilities must sum up
to 1.
k+1
3+1
2= 1 =⇒k= 1 −1
3−1
2=1
6
Step 2: Calculate the expected value of Xusing the formula:
E(X) = X
i
Xi·P(Xi)
Substitute the values of Xand P(X) into the formula:
E(X)=(−2) ·1
6+ (1) ·1
3+ (3) ·1
2
Calculate the expected value:
E(X) = −2
6+1
3+3
2=−1
3+1
3+3
2=5
2
Therefore, the value of kis 1
6and the expected value of Xis 5
2.
Question 17
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 2) = 0.2, P (X= 3) = 0.3, P (X= 5) = 0.5
Calculate the expected value E(X) of X.
12
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)=2·0.2+3·0.3+5·0.5
Step 3: Calculate the expected value:
E(X)=0.4+0.9+2.5=3.8
Step 4: Therefore, the expected value of the random variable Xis E(X) =
3.8.
Question 18
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.
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: Calculate the expected value by substituting the given probabilities:
E(X)=(−2) ·1
8+ 0 ·3
8+ 2 ·1
4+ 4 ·1
4.
Step 3: Simplify the expression:
E(X) = −2
8+0+2
4+4
4=−1
4+6
4=5
4.
Therefore, the expected value of the random variable Xis 5
4.
13
Question 19
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
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
i
xi·P(X=xi),
where the sum is taken over all possible values of X.
Step 2: In this case, we have:
E(X)=1·1
2+ 2 ·1
4+ 3 ·1
8+ 4 ·1
8.
Step 3: Simplifying the expression, we get:
E(X) = 1
2+1
2+3
8+1
2=13
4.
Step 4: Therefore, the expected value of Xis 13
4.
Question 20
Question
Let Xbe a discrete random variable with the following probability distribution:
X P (X)
−2 0.1
0 0.4
1 0.3
2 0.2
Find the expected value of X.
14
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 distribution of Xinto the
formula:
E(X) = (−2) ·0.1+0·0.4+1·0.3+2·0.2
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.3+0.4 = 0.5
Therefore, the expected value of the random variable Xis 0.5 .
Question 21
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.2, P (X= 0) = 0.4, P (X= 1) = 0.3, P (X= 2) = 0.1
Find 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: In this case, we have:
E(X) = (−1) ·0.2+0·0.4+1·0.3+2·0.1
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.3+0.2 = 0.3
Therefore, the expected value of Xis 0.3.
15
Question 22
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
X0 1 2 3
P(X=x) 0.1 0.3 0.4 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)
Step 2: First, we calculate the product of each xand its corresponding
probability P(X=x):
0·0.1+1·0.3+2·0.4+3·0.2
Step 3: Simplifying the expression above, we get:
0+0.3+0.8+0.6
Step 4: Adding these values together, we find:
E(X) = 0 + 0.3+0.8+0.6=1.7
Therefore, the expected value of the random variable Xis 1.7 .
Question 23
Question
Let Xbe a discrete random variable with the following probability distribution:
x−1 0 2
P(X=x)1
4
1
2
1
4
Calculate the expected value of X.
16
Solution
Step 1: First, we recall the definition of the expected value of a discrete random
variable. The expected value of X, denoted by E(X), is given by:
E(X) = X
all x
x·P(X=x)
Step 2: Using the probability distribution given, we can calculate the ex-
pected value E(X) as follows:
E(X)=(−1) ·1
4+ (0) ·1
2+ (2) ·1
4
Step 3: Simplifying the above expression, we get:
E(X) = −1
4+0+1
2
Step 4: Therefore, the expected value of Xis:
E(X) = 1
4
Thus, the expected value of Xis 1
4.
Question 24
Question
Let Xbe a discrete random variable with the following probability distribution:
x1 2 3
P(X=x)1
2
1
3
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 values of xand P(X=x) into the formula:
E(X)=1·1
2+ 2 ·1
3+ 3 ·1
6
Step 3: Simplify the expression:
17
E(X) = 1
2+2
3+1
2
Step 4: Find a common denominator and add the fractions:
E(X) = 3
6+4
6+1
6
Step 5: Combine the fractions:
E(X) = 8
6
Step 6: Simplify the fraction:
E(X) = 4
3
Therefore, the expected value of the random variable Xis 4
3.
Question 25
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
4, P (X= 0) = 1
2,and 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 given
by
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values xof X.
Step 2: In this case, Xcan take the values -1, 0, and 1. So, we have
E(X)=(−1) ·1
4+ (0) ·1
2+ (1) ·1
4
Step 3: Simplify the expression by multiplying each value of Xwith its
corresponding probability and summing them up.
E(X) = −1
4+0+1
4= 0
Step 4: Therefore, the expected value of the discrete random variable Xis
0 .
18
Question 26
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 2) = 0.2, P (X= 3) = 0.3, P (X= 4) = 0.4
Calculate the expected value of X.
Solution
Step 1: Calculate the expected value of Xusing the formula:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the given probabilities into the formula:
E(X) = (−1) ·0.1 + (2) ·0.2 + (3) ·0.3 + (4) ·0.4
Step 3: Calculate the expected value:
E(X) = −0.1+0.4+0.9+1.6=2.8
Therefore, the expected value of Xis 2.8.
Question 27
Question
Let Xbe a discrete random variable with the following probability distribution:
X023
P(X) 0.4 0.3 0.3
Calculate 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
x
x·P(X=x)
Step 2: Applying the formula for expected value, we have:
E(X)=0·0.4+2·0.3+3·0.3
19
Step 3: Calculate the expected value:
E(X) = 0 + 0.6+0.9
Step 4: Simplify to find the expected value:
E(X)=1.5
Therefore, the expected value of the discrete random variable Xis 1.5.
Question 28
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−3) = 1
12, P (X=−2) = 1
4, P (X= 0) = 1
6, P (X= 1) = 1
3, P (X= 2) = 1
4.
Find the expected value of X.
Solution
Step 1: First, find the expected value formula for a discrete random variable.
The expected value (or mean) of a discrete random variable Xis given by the
formula:
E(X) = Xxi·P(X=xi),
where the sum is taken over all possible values xithat Xcan take on.
Step 2: Compute the expected value of Xusing the probability mass function
provided.
E(X)=(−3)·P(X=−3)+(−2)·P(X=−2)+(0)·P(X= 0)+(1)·P(X= 1)+(2)·P(X= 2) = −3·1
12+−2·1
4+0·1
6+1·1
3+2·1
4.
Step 3: Calculate the expected value.
E(X) = −1
4−1
2+0+1
3+1
2=−1
4−1
2+2
6+3
6+4
6=8
12 =2
3.
Therefore, the expected value of the random variable Xis 2
3.
20
Question 29
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = (2
3k+1 if k= 0,1,2, . . .
0 otherwise
Find the expected value of X.
Solution
Step 1: First, we need to find the expected value of X, denoted by E[X]. The
formula for finding the expected value of a discrete random variable is given by:
E[X] = X
all k
k·P(X=k)
Step 2: We substitute the given probability mass function into the formula:
E[X] =
∞
X
k=0
k·2
3k+1
Step 3: We can simplify the expression before evaluating the sum:
E[X]=2
∞
X
k=0
k
3k+1
Step 4: We recognize that the sum above is in the form of the derivative of
a power series. Let’s differentiate P∞
k=0 1
3kwith respect to x:
d
dx ∞
X
k=0 1
3k!=
∞
X
k=0
k1
3k−1
Step 5: Simplifying the derivative gives us:
∞
X
k=0
k1
3k−1
=1
(1 −1
3)2=9
4
Step 6: So, we have found that P∞
k=0 k·1
3k=9
4. Therefore, the expected
value of the random variable Xis:
E[X]=2·9
4=9
2
Thus, the expected value of Xis 9
2.
21
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 1 3
P(X=x)1
6
2
3
1
6
Calculate the expected value of X.
Solution
Step 1: Identify the random variable and its probability distribution.
Given Xwith probability distribution:
x−2 1 3
P(X=x)1
6
2
3
1
6
Step 2: Calculate the expected value using the formula:
E(X) = X
x
x·P(X=x)
Step 3: Substitute the values from the probability distribution into the for-
mula and calculate the expected value.
E(X)=(−2) 1
6+ (1) 2
3+ (3) 1
6
E(X) = −2
6+2
3+3
6
Step 4: Simplify the expression to find the expected value of X.
E(X) = −1
3+2
3+1
2
E(X) = 1
2
Therefore, the expected value of Xis 1
2.
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
22
P(X=−2) = 1
8, P (X= 0) = 3
8, P (X= 1) = 1
4, P (X= 2) = 1
4
Calculate the expected value of X.
Solution
Given the probability mass function of X, we can calculate the expected value
of Xusing the formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 1: Identify the values and probabilities The values of Xare −2,
0, 1, and 2 with corresponding probabilities 1
8,3
8,1
4, and 1
4respectively.
Step 2: Calculate the expected value
E(X)=(−2) ·1
8+ (0) ·3
8+ (1) ·1
4+ (2) ·1
4
E(X) = −2
8+0+1
4+2
4
E(X) = −1
4+1
4+1
2
E(X) = 1
2
Therefore, the expected value of the random variable Xis 1
2.
Question 32
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= 4) = 0.3, P (X= 6) = 0.2
Find the expected value of X.
23
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) = (−2) ·0.1 + (0) ·0.4 + (4) ·0.3 + (6) ·0.2
Step 3: Calculate the expected value:
E(X) = −0.2+0+1.2+1.2=2.2
Therefore, the expected value of the random variable Xis 2.2.
Question 33
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
16, P (X= 4) = 1
32, P (X= 5) = 1
64,and P(X= 6) = 1
128
Calculate the expected value of X.
Solution
Step 1: Calculate the expected value of a discrete random variable X, denoted
as E[X], using the formula:
E[X] = Xxi·P(X=xi)
where xiare the possible values of X.
Step 2: Substitute the given probabilities into the formula for expected value:
E[X]=1·1
4+ 2 ·1
8+ 3 ·1
16 + 4 ·1
32 + 5 ·1
64 + 6 ·1
128
Step 3: Simplify the expression:
E[X] = 1
4+2
8+3
16 +4
32 +5
64 +6
128
E[X] = 32 + 32 + 24 + 16 + 10 + 6
128
24
E[X] = 120
128 =15
16
Therefore, the expected value of the discrete random variable Xis 15
16 .
Question 34
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−2) = 1
6, P (X= 0) = 2
3, P (X= 3) = 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
i
xi·P(X=xi)
where the sum is taken over all possible values of X(in this case, -2, 0, and
3).
Step 2: Calculate the expected value E(X) using the formula:
E(X)=(−2) ·1
6+ (0) ·2
3+ (3) ·1
6
Step 3: Simplify the expression to find the expected value E(X):
E(X) = −2
6+0+3
6=1
3
Therefore, the expected value of the discrete random variable Xis 1
3.
Question 35
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 2) = 1
4, P (X= 3) = 1
2.
Find the expected value of X.
25
Question 2
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.2, P (X= 2) = 0.3, P (X= 3) = 0.1, P (X= 4) = 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
x
x·P(X=x),
where the sum is taken over all possible values of X.
Step 2: Calculate the expected value E(X) using the given probability mass
function:
E(X) = 1 ·0.2+2·0.3+3·0.1+4·0.4.
Step 3: Compute the expected value of X:
E(X)=0.2+0.6+0.3+1.6=2.7.
Therefore, the expected value of the random variable Xis 2.7.
Question 3
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
0.2 if x= 1,
0.3 if x= 2,
0.1 if x= 3,
0.2 if x= 4,
0.2 if x= 5.
Calculate the expected value of X.
2
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=x) into the formula:
E(X)=1·0.2+2·0.3+3·0.1+4·0.2+5·0.2
Step 3: Perform the calculations:
E(X)=0.2+0.6+0.3+0.8 + 1 = 3
Step 4: Therefore, the expected value of the random variable Xis 3.
Question 4
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= 1) = 0.2, P (X= 2) = 0.4
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 values from the provided probability mass function
into the formula:
E(X)=(−2) ·0.1 + (0) ·0.3 + (1) ·0.2 + (2) ·0.4
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.2+0.8 = 0.8
Therefore, the expected value of the random variable Xis 0.8.
3
Question 5
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 3) = 1
2, P (X= 5) = 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) = Xxi·P(X=xi)
where the sum is taken over all possible values xiof X.
Step 2: In this case, we have X={1,3,5}and their corresponding proba-
bilities:
P(X= 1) = 1
4, P (X= 3) = 1
2, P (X= 5) = 1
4
Step 3: Now, we can calculate the expected value using the formula:
E(X) = 1 ·1
4+ 3 ·1
2+ 5 ·1
4
Step 4: Simplifying the expression, we get:
E(X) = 1
4+3
2+5
4
E(X) = 1
4+6
4+5
4
E(X) = 12
4
Step 5: Finally, we have:
E(X) = 12
4= 3
Therefore, the expected value of the random variable Xis 3.
Question 6
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.2, P (X= 2) = 0.3, P (X= 3) = 0.4, P (X= 4) = 0.1.
Find the expected value of X.
4
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 xthat Xcan take on.
Step 2: Substitute the given probabilities into the formula for expected value:
E(X) = 1 ·0.2+2·0.3+3·0.4+4·0.1.
Step 3: Calculate the expected value:
E(X)=0.2+0.6+1.2+0.4=2.4.
Therefore, the expected value of the random variable Xis 2.4 .
Question 7
Question
Let Xbe a discrete random variable with the following probability distribution:
X123
P(X) 0.3 0.5 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: We can now substitute the values from the probability distribution
into the formula for expected value:
E(X)=1·0.3+2·0.5+3·0.2
Step 3: Calculate the products and sum them up:
E(X) = 0.3+1.0+0.6
Step 4: Finally, sum up the values to get the expected value of X:
E(X)=1.9
Therefore, the expected value of the random variable Xis 1.9.
5
Question 8
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.2, P (X= 0) = 0.5, P (X= 1) = 0.3.
Calculate the expected value of X.
Solution
Step 1: Find the expected value of Xusing 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:
E(X)=(−1)(0.2) + (0)(0.5) + (1)(0.3).
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.3 = 0.1.
Therefore, the expected value of random variable Xis 0.1.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.25, P (X= 1) = 0.1, P (X= 3) = 0.2, P (X= 4) = 0.45.
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 of X.
6
Step 2: Calculate the expected value E(X) using the given probability mass
function:
E(X)=(−2) ·0.25 + (1) ·0.1 + (3) ·0.2 + (4) ·0.45
Step 3: Multiply each value by its probability and sum them up:
E(X) = −0.5+0.1+0.6+1.8
Step 4: Calculate the final expected value:
E(X)=2.0
Question 10
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 3
P(X) 0.2 0.5 0.3
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 of Xand P(X=xi) are the corresponding
probabilities.
Step 2: Substitute the values from the probability distribution into the for-
mula:
E(X)=(−2) ·0.2 + (0) ·0.5 + (3) ·0.3
Step 3: Perform the calculations:
E(X) = −0.4+0+0.9
E(X)=0.5
Therefore, the expected value of the discrete random variable Xis 0.5.
7
Question 11
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 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 from the probability mass function into the
formula:
E(X)=1·1
3+ 2 ·1
6+ 3 ·1
2
Step 3: Simplify the expression:
E(X) = 1
3+1
3+3
2
Step 4: Combine the fractions:
E(X) = 2
3+3
2=4
6+9
6=13
6
Step 5: Therefore, the expected value of the random variable Xis 13
6.
Question 12
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= 1) = 0.2, P (X= 3) = 0.4.
Find the expected value of X.
8
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: Calculate the expected value by multiplying each value of Xby its
corresponding probability and summing them up:
E(X)=(−2)(0.1) + (0)(0.3) + (1)(0.2) + (3)(0.4)
E(X) = −0.2+0+0.2+1.2
E(X) = 1.2
Therefore, the expected value of the random variable Xis 1.2.
Question 13
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2 3 4
P(X)1
4
1
4
1
4
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
x
x·P(X=x)
Step 2: Substitute the values of xand P(X=x) from the probability
distribution into the formula:
E(X)=1·1
4+ 2 ·1
4+ 3 ·1
4+ 4 ·1
4
Step 3: Simplify the expression:
E(X) = 1
4+2
4+3
4+4
4
E(X) = 10
4
9
E(X)=2.5
Step 4: Therefore, the expected value of the random variable Xis 2.5.
Question 14
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2 3
P(X) 0.2 0.3p
If E[X] = 2.2, find the value of p.
Solution
Step 1: The expected value E[X] 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 xthat Xcan take.
Given that E[X]=2.2 and the probability distribution of X, we can write:
E[X]=1·0.2+2·0.3+3·p
Step 2: Simplify the equation:
2.2=0.2+0.6+3p
2.2=0.8+3p
1.4=3p
p=1.4
3= 0.4667
Therefore, the value of pis 0.4667 .
10
Question 15
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,and P(X=k) = 0.4 for some k∈Z.
Find the value of kand calculate the expected value of X.
Solution
Step 1: Find the value of k Since the sum of all probabilities must equal 1,
we have:
0.1+0.2+0.3+0.4 = 1 =⇒1.0 = 1
Therefore, the missing probability is:
0.4 = P(X=k)
0.4=1−0.1−0.2−0.3
0.4=0.4
Hence, kmust be equal to 1.
Step 2: Calculate the expected value of X The expected value of a
discrete random variable Xis given by:
E(X) = Xx·P(X=x)
Substitute the values from the probability mass function:
E(X)=(−2)(0.1) + (0)(0.2) + (3)(0.3) + (1)(0.4)
E(X) = −0.2+0+0.9+0.4
E(X)=1.1
Therefore, the expected value of Xis 1.
11
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 1 3
P(X)k1
3
1
2
Find the value of kand calculate the expected value of X.
Solution
Step 1: Find the value of kusing the fact that the probabilities must sum up
to 1.
k+1
3+1
2= 1 =⇒k= 1 −1
3−1
2=1
6
Step 2: Calculate the expected value of Xusing the formula:
E(X) = X
i
Xi·P(Xi)
Substitute the values of Xand P(X) into the formula:
E(X)=(−2) ·1
6+ (1) ·1
3+ (3) ·1
2
Calculate the expected value:
E(X) = −2
6+1
3+3
2=−1
3+1
3+3
2=5
2
Therefore, the value of kis 1
6and the expected value of Xis 5
2.
Question 17
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 2) = 0.2, P (X= 3) = 0.3, P (X= 5) = 0.5
Calculate the expected value E(X) of X.
12
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)=2·0.2+3·0.3+5·0.5
Step 3: Calculate the expected value:
E(X)=0.4+0.9+2.5=3.8
Step 4: Therefore, the expected value of the random variable Xis E(X) =
3.8.
Question 18
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.
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: Calculate the expected value by substituting the given probabilities:
E(X)=(−2) ·1
8+ 0 ·3
8+ 2 ·1
4+ 4 ·1
4.
Step 3: Simplify the expression:
E(X) = −2
8+0+2
4+4
4=−1
4+6
4=5
4.
Therefore, the expected value of the random variable Xis 5
4.
13
Question 19
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
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
i
xi·P(X=xi),
where the sum is taken over all possible values of X.
Step 2: In this case, we have:
E(X)=1·1
2+ 2 ·1
4+ 3 ·1
8+ 4 ·1
8.
Step 3: Simplifying the expression, we get:
E(X) = 1
2+1
2+3
8+1
2=13
4.
Step 4: Therefore, the expected value of Xis 13
4.
Question 20
Question
Let Xbe a discrete random variable with the following probability distribution:
X P (X)
−2 0.1
0 0.4
1 0.3
2 0.2
Find the expected value of X.
14
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 distribution of Xinto the
formula:
E(X) = (−2) ·0.1+0·0.4+1·0.3+2·0.2
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.3+0.4 = 0.5
Therefore, the expected value of the random variable Xis 0.5 .
Question 21
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.2, P (X= 0) = 0.4, P (X= 1) = 0.3, P (X= 2) = 0.1
Find 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: In this case, we have:
E(X) = (−1) ·0.2+0·0.4+1·0.3+2·0.1
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.3+0.2 = 0.3
Therefore, the expected value of Xis 0.3.
15
Question 22
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
X0 1 2 3
P(X=x) 0.1 0.3 0.4 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)
Step 2: First, we calculate the product of each xand its corresponding
probability P(X=x):
0·0.1+1·0.3+2·0.4+3·0.2
Step 3: Simplifying the expression above, we get:
0+0.3+0.8+0.6
Step 4: Adding these values together, we find:
E(X) = 0 + 0.3+0.8+0.6=1.7
Therefore, the expected value of the random variable Xis 1.7 .
Question 23
Question
Let Xbe a discrete random variable with the following probability distribution:
x−1 0 2
P(X=x)1
4
1
2
1
4
Calculate the expected value of X.
16
Solution
Step 1: First, we recall the definition of the expected value of a discrete random
variable. The expected value of X, denoted by E(X), is given by:
E(X) = X
all x
x·P(X=x)
Step 2: Using the probability distribution given, we can calculate the ex-
pected value E(X) as follows:
E(X)=(−1) ·1
4+ (0) ·1
2+ (2) ·1
4
Step 3: Simplifying the above expression, we get:
E(X) = −1
4+0+1
2
Step 4: Therefore, the expected value of Xis:
E(X) = 1
4
Thus, the expected value of Xis 1
4.
Question 24
Question
Let Xbe a discrete random variable with the following probability distribution:
x1 2 3
P(X=x)1
2
1
3
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 values of xand P(X=x) into the formula:
E(X)=1·1
2+ 2 ·1
3+ 3 ·1
6
Step 3: Simplify the expression:
17
E(X) = 1
2+2
3+1
2
Step 4: Find a common denominator and add the fractions:
E(X) = 3
6+4
6+1
6
Step 5: Combine the fractions:
E(X) = 8
6
Step 6: Simplify the fraction:
E(X) = 4
3
Therefore, the expected value of the random variable Xis 4
3.
Question 25
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
4, P (X= 0) = 1
2,and 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 given
by
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values xof X.
Step 2: In this case, Xcan take the values -1, 0, and 1. So, we have
E(X)=(−1) ·1
4+ (0) ·1
2+ (1) ·1
4
Step 3: Simplify the expression by multiplying each value of Xwith its
corresponding probability and summing them up.
E(X) = −1
4+0+1
4= 0
Step 4: Therefore, the expected value of the discrete random variable Xis
0 .
18
Question 26
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 2) = 0.2, P (X= 3) = 0.3, P (X= 4) = 0.4
Calculate the expected value of X.
Solution
Step 1: Calculate the expected value of Xusing the formula:
E(X) = X
x
x·P(X=x)
Step 2: Substitute the given probabilities into the formula:
E(X) = (−1) ·0.1 + (2) ·0.2 + (3) ·0.3 + (4) ·0.4
Step 3: Calculate the expected value:
E(X) = −0.1+0.4+0.9+1.6=2.8
Therefore, the expected value of Xis 2.8.
Question 27
Question
Let Xbe a discrete random variable with the following probability distribution:
X023
P(X) 0.4 0.3 0.3
Calculate 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
x
x·P(X=x)
Step 2: Applying the formula for expected value, we have:
E(X)=0·0.4+2·0.3+3·0.3
19
Step 3: Calculate the expected value:
E(X) = 0 + 0.6+0.9
Step 4: Simplify to find the expected value:
E(X)=1.5
Therefore, the expected value of the discrete random variable Xis 1.5.
Question 28
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−3) = 1
12, P (X=−2) = 1
4, P (X= 0) = 1
6, P (X= 1) = 1
3, P (X= 2) = 1
4.
Find the expected value of X.
Solution
Step 1: First, find the expected value formula for a discrete random variable.
The expected value (or mean) of a discrete random variable Xis given by the
formula:
E(X) = Xxi·P(X=xi),
where the sum is taken over all possible values xithat Xcan take on.
Step 2: Compute the expected value of Xusing the probability mass function
provided.
E(X)=(−3)·P(X=−3)+(−2)·P(X=−2)+(0)·P(X= 0)+(1)·P(X= 1)+(2)·P(X= 2) = −3·1
12+−2·1
4+0·1
6+1·1
3+2·1
4.
Step 3: Calculate the expected value.
E(X) = −1
4−1
2+0+1
3+1
2=−1
4−1
2+2
6+3
6+4
6=8
12 =2
3.
Therefore, the expected value of the random variable Xis 2
3.
20
Question 29
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) = (2
3k+1 if k= 0,1,2, . . .
0 otherwise
Find the expected value of X.
Solution
Step 1: First, we need to find the expected value of X, denoted by E[X]. The
formula for finding the expected value of a discrete random variable is given by:
E[X] = X
all k
k·P(X=k)
Step 2: We substitute the given probability mass function into the formula:
E[X] =
∞
X
k=0
k·2
3k+1
Step 3: We can simplify the expression before evaluating the sum:
E[X]=2
∞
X
k=0
k
3k+1
Step 4: We recognize that the sum above is in the form of the derivative of
a power series. Let’s differentiate P∞
k=0 1
3kwith respect to x:
d
dx ∞
X
k=0 1
3k!=
∞
X
k=0
k1
3k−1
Step 5: Simplifying the derivative gives us:
∞
X
k=0
k1
3k−1
=1
(1 −1
3)2=9
4
Step 6: So, we have found that P∞
k=0 k·1
3k=9
4. Therefore, the expected
value of the random variable Xis:
E[X]=2·9
4=9
2
Thus, the expected value of Xis 9
2.
21
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 1 3
P(X=x)1
6
2
3
1
6
Calculate the expected value of X.
Solution
Step 1: Identify the random variable and its probability distribution.
Given Xwith probability distribution:
x−2 1 3
P(X=x)1
6
2
3
1
6
Step 2: Calculate the expected value using the formula:
E(X) = X
x
x·P(X=x)
Step 3: Substitute the values from the probability distribution into the for-
mula and calculate the expected value.
E(X)=(−2) 1
6+ (1) 2
3+ (3) 1
6
E(X) = −2
6+2
3+3
6
Step 4: Simplify the expression to find the expected value of X.
E(X) = −1
3+2
3+1
2
E(X) = 1
2
Therefore, the expected value of Xis 1
2.
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
22
P(X=−2) = 1
8, P (X= 0) = 3
8, P (X= 1) = 1
4, P (X= 2) = 1
4
Calculate the expected value of X.
Solution
Given the probability mass function of X, we can calculate the expected value
of Xusing the formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 1: Identify the values and probabilities The values of Xare −2,
0, 1, and 2 with corresponding probabilities 1
8,3
8,1
4, and 1
4respectively.
Step 2: Calculate the expected value
E(X)=(−2) ·1
8+ (0) ·3
8+ (1) ·1
4+ (2) ·1
4
E(X) = −2
8+0+1
4+2
4
E(X) = −1
4+1
4+1
2
E(X) = 1
2
Therefore, the expected value of the random variable Xis 1
2.
Question 32
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= 4) = 0.3, P (X= 6) = 0.2
Find the expected value of X.
23
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) = (−2) ·0.1 + (0) ·0.4 + (4) ·0.3 + (6) ·0.2
Step 3: Calculate the expected value:
E(X) = −0.2+0+1.2+1.2=2.2
Therefore, the expected value of the random variable Xis 2.2.
Question 33
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
16, P (X= 4) = 1
32, P (X= 5) = 1
64,and P(X= 6) = 1
128
Calculate the expected value of X.
Solution
Step 1: Calculate the expected value of a discrete random variable X, denoted
as E[X], using the formula:
E[X] = Xxi·P(X=xi)
where xiare the possible values of X.
Step 2: Substitute the given probabilities into the formula for expected value:
E[X]=1·1
4+ 2 ·1
8+ 3 ·1
16 + 4 ·1
32 + 5 ·1
64 + 6 ·1
128
Step 3: Simplify the expression:
E[X] = 1
4+2
8+3
16 +4
32 +5
64 +6
128
E[X] = 32 + 32 + 24 + 16 + 10 + 6
128
24
E[X] = 120
128 =15
16
Therefore, the expected value of the discrete random variable Xis 15
16 .
Question 34
Question
Let Xbe a discrete random variable with probability mass function given by:
P(X=−2) = 1
6, P (X= 0) = 2
3, P (X= 3) = 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
i
xi·P(X=xi)
where the sum is taken over all possible values of X(in this case, -2, 0, and
3).
Step 2: Calculate the expected value E(X) using the formula:
E(X)=(−2) ·1
6+ (0) ·2
3+ (3) ·1
6
Step 3: Simplify the expression to find the expected value E(X):
E(X) = −2
6+0+3
6=1
3
Therefore, the expected value of the discrete random variable Xis 1
3.
Question 35
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 1
4, P (X= 2) = 1
4, P (X= 3) = 1
2.
Find the expected value of X.
25
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
4+ 2 ·1
4+ 3 ·1
2.
Step 3: Simplify the expression:
E(X) = 1
4+2
4+3
2=1+2+6
4=9
4= 2.25 .
Therefore, the expected value of the random variable Xis 2.25.
26
