MATH 350 - DISCRETE
MATHEMATICS - Discrete random
variables and expected value
Question Bank - Set 3
Liberty University
Question 1
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−1) = 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: 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:
E(X)=(−1)(0.1) + (0)(0.3) + (1)(0.2) + (2)(0.4)
Step 3: Calculate the expected value:
E(X) = −0.1+0+0.2+0.8
E(X)=0.9
Therefore, the expected value of the random variable Xis 0.9.
Question 2
Question
Let X be a discrete random variable with the following probability distribution:
x−1 0 2
P(X=x)1
3
1
6
1
2
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X is given
by:
E(X) = X
all x
x·P(X=x)
Step 2: Calculate the expected value E(X) using the probability distribution
given:
E(X)=(−1) ·1
3+ (0) ·1
6+ (2) ·1
2
Step 3: Perform the calculations:
E(X) = −1
3+ 0 + 1 = 2
3
Step 4: Therefore, the expected value of the random variable X is 2
3.
Question 3
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.3, P (X= 2) = 0.4, P (X= 3) = 0.2, P (X= 4) = 0.1
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 values of xand P(X=x) into the formula:
2
E(X)=1·0.3+2·0.4+3·0.2+4·0.1
Step 3: Simplify the expression:
E(X)=0.3+0.8+0.6+0.4
E(X)=2.1
Therefore, the expected value of the random variable Xis 2.1.
Question 4
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) = 0.6
Compute 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: Calculate the expected value by substituting the given probabilities
into the formula:
E(X)=(−1) ·0.1+0·0.3+1·0.6
Step 3: Simplify the expression:
E(X) = −0.1+0+0.6=0.5
Step 4: Therefore, the expected value of the random variable Xis 0.5 .
Question 5
Question
Let Xbe a discrete random variable with the following probability mass function
(pmf):
3
P(X=−3) = 0.1, P (X= 0) = 0.2, P (X= 1) = 0.4, P (X= 2) = 0.3.
Calculate the expected value of the random variable X.
Solution
Step 1: The expected value of a discrete random variable is denoted by E[X]
and is calculated by summing the product of each possible value of the random
variable and its corresponding probability:
E[X] = X
x
x·P(X=x).
Step 2: Given the pmf for random variable X, we can calculate the expected
value as follows:
E[X] = (−3)(0.1) + (0)(0.2) + (1)(0.4) + (2)(0.3).
Step 3: Simplifying the expression yields:
E[X] = −0.3+0+0.4+0.6.
Step 4: Therefore, the expected value of the random variable Xis:
E[X]=0.7.
Question 6
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2 3 4
P(X=x)1
8
1
4
1
201
8
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:
4
E(X) = 0 ·1
8+ 1 ·1
4+ 2 ·1
2+ 3 ·0+4·1
8
Step 3: Simplify the expression:
E(X) = 0 + 1
4+1+0+1
2
E(X) = 1
4+2
2+1
2
Step 4: Combine the fractions:
E(X) = 1
4+2
2+1
2=1
4+1+1
2=1
4+2
4+2
4=5
4
Step 5: Therefore, the expected value of Xis 5
4.
Question 7
Question
Let Xbe a discrete random variable with probability mass function given by
P(X=−2) = 0.1, P (X= 0) = 0.4, P (X= 1) = 0.3, P (X= 3) = 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
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 and calculate the
expected value.
E(X) = (−2) ·0.1 + (0) ·0.4 + (1) ·0.3 + (3) ·0.2
E(X) = −0.2+0+0.3+0.6
E(X) = 0.7.
Therefore, the expected value of the random variable Xis 0.7.
5
Question 8
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
8, P (X= 0) = 1
4, P (X= 1) = 1
2, P (X= 2) = 1
8
Find 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 given values into the formula to find E(X):
E(X)=(−2) ·1
8+ (0) ·1
4+ (1) ·1
2+ (2) ·1
8
Step 3: Simplify the expression:
E(X) = −2
8+0+1
2+2
8=−2
8+4
8+1
2+2
8=5
8+4
8=9
8
Step 4: Therefore, the expected value of Xis 9
8.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
8, P (X= 0) = 2
8, P (X= 3) = 5
8
Find the expected value of X.
6
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
i
xi·P(X=xi)
Step 2: We will calculate E(X) using the given probability mass function.
Step 3: Substitute the values of xiand P(X=xi) into the formula:
E(X) = (−2) ·1
8+ 0 ·2
8+ 3 ·5
8
Step 4: Simplify the expression:
E(X) = −2
8+0+15
8=13
8
Step 5: Therefore, the expected value of the random variable Xis 13
8.
Question 10
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
0.1 for x= 1
0.2 for x= 2
0.3 for x= 3
0.4 for x= 4
0 otherwise
Find the expected value E[X] of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis cal-
culated as E[X] = Px·P(X=x) over all possible values of X.
Step 2: In this case, we have the PMF P(X=x) defined for x= 1,2,3,4.
Step 3: We can now calculate the expected value E[X] by multiplying each
possible value of Xby its corresponding probability and summing the results:
E[X] = X
x
x·P(X=x)=1·0.1+2·0.2+3·0.3+4·0.4
Step 4: Now, we can compute the expected value:
E[X]=0.1+0.4+0.9+1.6=3
Therefore, the expected value of the random variable Xis E[X] = 3.
7
Question 11
Question
Let Xbe a discrete random variable with the following probability distribution:
x012
P(X=x) 0.2 0.5 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
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) are their respective
probabilities.
Step 2: Calculate the expected value E(X) using the provided probability
distribution:
E(X)=0·0.2+1·0.5+2·0.3
Step 3: Simplify the expression:
E(X) = 0 + 0.5+0.6
E(X)=1.1
Step 4: Therefore, the expected value of the random variable Xis E(X) =
1.1.
Question 12
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
9, P (X= 0) = 2
9, P (X= 3) = 2
9, P (X= 5) = 4
9
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) = Xx·P(X=x)
Step 2: Calculate the expected value E(X) by summing the products of each
value of Xwith its corresponding probability:
E(X)=(−2) ·1
9+ 0 ·2
9+ 3 ·2
9+ 5 ·4
9
Step 3: Simplify the expression:
E(X) = −2
9+0+6
9+20
9=24
9
Step 4: Reduce the fraction:
E(X) = 8
3
Therefore, the expected value of the random variable Xis 8
3.
Question 13
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) = 0.4, P (X= 3) = 0.2.
Calculate the expected value of X.
Solution
Step 1: The expected value, denoted by µ, of a discrete random variable Xis
calculated as:
E[X] = X
x
x·P(X=x),
where the sum is taken over all possible values xthat Xcan take.
Step 2: Substituting the values from the probability mass function into the
formula:
E[X] = (−2) ·0.1 + (0) ·0.3 + (1) ·0.4 + (3) ·0.2
9
Step 3: Calculating the expected value:
E[X] = −0.2+0+0.4+0.6 = 0.8.
Therefore, the expected value of the random variable Xis 0.8.
Question 14
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= 2) = 0.3, P (X= 4) = 0.2
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 of X.
Step 2: Substitute the given probabilities into the formula:
E(X)=(−2)(0.1) + (0)(0.4) + (2)(0.3) + (4)(0.2)
Step 3: Perform the calculations:
E(X) = −0.2+0+0.6+0.8 = 1.2
Step 4: Therefore, the expected value of the random variable Xis 1.2 .
Question 15
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2
P(X=x) 0.3 0.4 0.3
Calculate the expected value of X.
10
Solution
The expected value of a discrete random variable Xis given by:
E(X) = X
all x
x·P(X=x)
Step 1: Calculate the expected value of Xusing the probability distribution
provided.
E(X)=0·0.3+1·0.4+2·0.3
= 0 + 0.4+0.6
= 1
Step 2: Therefore, the expected value of the random variable Xis 1.
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 1 3
P(X) 0.2 0.3a0.1
If E[X] = 0.6, find the value of a.
Solution
Step 1: Recall that the expected value E[X] of a discrete random variable Xis
given by the formula:
E[X] = X
x
x·P(X=x)
Step 2: Substitute the given values into the formula for E[X] and solve for
E[X]:
E[X]=(−2)(0.2) + (0)(0.3) + (1)(a) + (3)(0.1)
0.6 = −0.4 + 0 + a+ 0.3
Step 3: Simplify the equation:
0.6 = −0.1 + a
a= 0.6+0.1=0.7
Step 4: Therefore, the value of ais 0.7.
11
Question 17
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
0.4 if x= 4
0 otherwise
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: Calculate the expected value of Xusing the probability mass func-
tion provided:
E(X) = (1)(0.2) + (2)(0.3) + (3)(0.1) + (4)(0.4)
Step 3: Simplify the expression by multiplying the values of xby their
respective probabilities:
E(X)=0.2+0.6+0.3+1.6
Step 4: Add up the terms to find the expected value of X:
E(X)=2.7
Therefore, the expected value of the random variable Xis 2.7.
Question 18
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
5, P (X= 0) = 2
5, P (X= 1) = 2
5
Calculate the expected value of X.
12
Solution
Step 1: First, 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: Using the probability mass function provided, we can calculate the
expected value as follows:
E(X) = −1·1
5+ 0 ·2
5+ 1 ·2
5
Step 3: Simplifying the expression, we get:
E(X) = −1
5+0+2
5=1
5
Step 4: Therefore, the expected value of the random variable Xis 1
5.
Question 19
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) = 0.2, P (X= 2) = 0.4
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
as E(X), is calculated as:
E(X) = Xx·P(X=x)
Step 2: Substituting the given probability mass function values, we have:
E(X) = (−1) ·0.1 + (0) ·0.3 + (1) ·0.2 + (2) ·0.4
Step 3: Calculating the expected value:
E(X)=(−0.1) + (0) + (0.2) + (0.8)
E(X)=0.9
Therefore, the expected value of Xis 0.9.
13
Question 20
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.
Calculate the expected value of X.
Solution
To find the expected value of a discrete random variable X, denoted as E(X),
we use the formula:
E(X) = X
all x
x·P(X=x).
Step 1: Calculate the expected value E(X).
E(X)=1·0.2+2·0.3+3·0.4+4·0.1
Step 2: Simplify the expression.
E(X)=0.2+0.6+1.2+0.4
E(X)=2.4
Therefore, the expected value of the random variable Xis 2.4.
Question 21
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) =
0.2 if k= 1
0.3 if k= 2
0.1 if k= 3
0.4 if k= 4
Find the expected value of X.
14
Solution
1. First, recall that the expected value of a discrete random variable Xis
given by the formula:
E(X) = X
k
k·P(X=k)
where the sum is taken over all possible values of X.
2. Now, we can calculate the expected value of Xusing the given probability
mass function:
E(X) = X
k
k·P(X=k) = (1)(0.2) + (2)(0.3) + (3)(0.1) + (4)(0.4)
3. Now, we can plug in the values and simplify:
E(X) = 1(0.2) + 2(0.3) + 3(0.1) + 4(0.4) = 0.2+0.6+0.3+1.6 = 2.7
4. Therefore, the expected value of the random variable Xis 2.7.
Question 22
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.3, P (X= 0) = 0.4, P (X= 3) = 0.1, P (X=k)=0.2
Find the expected value of X.
Solution
Step 1: First, 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: We can now calculate the expected value of Xusing the probability
mass function provided.
E[X] = (−1)(0.3) + (0)(0.4) + (3)(0.1) + (k)(0.2)
Step 3: Simplifying, we get:
E[X] = −0.3+0+0.3+0.2k
15
Step 4: Further simplifying, we find:
E[X]=0.2k
Therefore, the expected value of the discrete random variable Xis 0.2k.
Question 23
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: The expected value of a discrete random variable Xis defined as:
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 for expected value:
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=3
Therefore, the expected value of the random variable Xis 3.
Question 24
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2 3 4
P(X) 0.1 0.3 0.4 0.2
Find the expected value of X.
16
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
by E(X), is given by 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: Calculate the expected value of Xusing the probability distribution
given:
E(X)=1·0.1+2·0.3+3·0.4+4·0.2
Step 3: Perform the calculations to find the expected value:
E(X) = 0.1+0.6+1.2+0.8=2.7
Step 4: Therefore, the expected value of the random variable Xis 2.7 .
Question 25
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
Calculate the expected value of X.
Solution
To find the expected value of a discrete random variable X, we use the formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 1: Calculate the expected value
The expected value E(X) of Xis given by:
E(X) = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
Substitute the given probabilities:
E(X) = 1 ·1
4+ 2 ·1
4+ 3 ·1
2
E(X) = 1
4+1
2+3
2
17
E(X) = 1
4+2
4+6
4
E(X) = 9
4
Therefore, the expected value of the random variable Xis 9
4.
Question 26
Question
Let Xbe a discrete random variable with the following probability distribution:
x P (X=x)x2P(X=x)
0 0.2 0
1 0.3 0.3
2 0.4 0.8
3 0.1 0.3
Calculate the expected value of X.
Solution
To find the expected value of X, denoted by E(X), we use the formula:
E(X) = X
x
x·P(X=x)
Given the probability distribution of X, we have:
E(X) = X
x
x·P(X=x)=0·0.2+1·0.3+2·0.4+3·0.1
Step 1: Calculate E(X)
E(X) = 0 ·0.2+1·0.3+2·0.4+3·0.1 = 0 + 0.3+0.8+0.3=1.4
Therefore, the expected value of Xis 1.4.
Question 27
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2
P(X=x) 0.2 0.5p
If E(X)=1.4, find the value of p.
18
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: In this case, we are given that E(X) = 1.4, so we can write the
expression for the expected value as:
E(X) = 0 ·0.2+1·0.5+2·p
Step 3: Now, we substitute E(X)=1.4 into the above equation:
1.4 = 0 + 0.5+2p
Step 4: Simplifying the equation, we get:
1.4 = 0.5+2p
Step 5: Subtracting 0.5 from both sides gives:
0.9=2p
Step 6: Finally, solving for pgives:
p=0.9
2= 0.45
Therefore, the value of pis 0.45.
Question 28
Question
Let Xbe a discrete random variable with the following probability distribution:
x−1 0 3
P(X=x)1
4
1
2
1
4
Find 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: Substitute the values of xand P(X=x) into the formula. We have:
19
E(X)=(−1) ·1
4+ 0 ·1
2+ 3 ·1
4
Step 3: Simplify the expression:
E(X) = −1
4+0+3
4
E(X) = 2
4
Step 4: Further simplify the fraction:
E(X) = 1
2
Therefore, the expected value of the random variable Xis 1
2.
Question 29
Question
Let Xbe a discrete random variable with the following probability distribution:
X0 1 2
P(X)1
3
1
2
1
6
Calculate the expected value of X, denoted as E[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 that Xcan take and P(X=xi) is the
probability of Xtaking the value xi.
Step 2: In this case, we have:
E[X]=0·1
3+ 1 ·1
2+ 2 ·1
6
Step 3: Simplify the expression:
E[X] = 0 + 1
2+1
3
Step 4: Calculate the sum:
E[X] = 1
2+1
3
20
Step 5: Find a common denominator:
E[X] = 3
6+2
6
Step 6: Add the fractions:
E[X] = 5
6
Therefore, the expected value of the random variable Xis E[X] = 5
6.
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 0 1
P(X=x) 0.3 0.4 0.3
Find 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: Calculate the expected value of Xusing the given probability dis-
tribution:
E(X)=(−2) ·0.3 + (0) ·0.4 + (1) ·0.3
Step 3: Simplify the expression:
E(X) = −0.6+0+0.3
Step 4: Calculate the final answer:
E(X) = −0.3
Therefore, the expected value of the random variable Xis −0.3 .
21
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 0) = 0.2, P (X= 1) = 0.3, 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 given probability mass function into the formula for
expected value:
E(X)=(−1)(0.1) + (0)(0.2) + (1)(0.3) + (2)(0.4)
Step 3: Calculate the expected value:
E(X) = −0.1+0+0.3+0.8
E(X) = 1
Therefore, the expected value of the random variable Xis 1.
Question 32
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.3, P (X= 3) = 0.6
Calculate the expected value of X.
22
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: Using the given probability mass function for X, we can calculate
the expected value as follows:
E(X)=(−2) ·0.1 + (0) ·0.3 + (3) ·0.6
Step 3: Simplifying, we get:
E(X) = −0.2+0+1.8
Step 4: Therefore, the expected value of Xis:
E(X)=1.6
Hence, the expected value of the random variable Xis 1.6.
Question 33
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
Find the expected value of X.
Solution
Step 1: The expected value of a discrete random variable X, denoted by E(X),
is defined by the formula:
E(X) = Xx·P(X=x)
Step 2: Given the probability mass function, we can calculate the expected
value of Xas follows:
E(X) = 1 ×0.1+2×0.2+3×0.3+4×0.4
Step 3: Simplifying the expression, we get:
23
E(X)=0.1+0.4+0.9+1.6
E(X) = 3
Step 4: Therefore, the expected value of the random variable Xis 3 .
Question 34
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−1) = 1
4, P (X= 0) = 1
2, P (X= 2) = 1
4
Calculate 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 defined as:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 2: In this case, the possible values of Xare -1, 0, and 2, with their
corresponding probabilities as given. We will substitute these values into the
formula for expected value and calculate the result:
E(X)=(−1) ·1
4+ (0) ·1
2+ (2) ·1
4
Step 3: Simplifying the expression, we get:
E(X) = −1
4+0+1
2
E(X) = 1
4
Therefore, the expected value of the random variable Xis 1
4.
24
Question 2
Question
Let X be a discrete random variable with the following probability distribution:
x−1 0 2
P(X=x)1
3
1
6
1
2
Calculate the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X is given
by:
E(X) = X
all x
x·P(X=x)
Step 2: Calculate the expected value E(X) using the probability distribution
given:
E(X)=(−1) ·1
3+ (0) ·1
6+ (2) ·1
2
Step 3: Perform the calculations:
E(X) = −1
3+ 0 + 1 = 2
3
Step 4: Therefore, the expected value of the random variable X is 2
3.
Question 3
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X= 1) = 0.3, P (X= 2) = 0.4, P (X= 3) = 0.2, P (X= 4) = 0.1
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 values of xand P(X=x) into the formula:
2
E(X)=1·0.3+2·0.4+3·0.2+4·0.1
Step 3: Simplify the expression:
E(X)=0.3+0.8+0.6+0.4
E(X)=2.1
Therefore, the expected value of the random variable Xis 2.1.
Question 4
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) = 0.6
Compute 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: Calculate the expected value by substituting the given probabilities
into the formula:
E(X)=(−1) ·0.1+0·0.3+1·0.6
Step 3: Simplify the expression:
E(X) = −0.1+0+0.6=0.5
Step 4: Therefore, the expected value of the random variable Xis 0.5 .
Question 5
Question
Let Xbe a discrete random variable with the following probability mass function
(pmf):
3
P(X=−3) = 0.1, P (X= 0) = 0.2, P (X= 1) = 0.4, P (X= 2) = 0.3.
Calculate the expected value of the random variable X.
Solution
Step 1: The expected value of a discrete random variable is denoted by E[X]
and is calculated by summing the product of each possible value of the random
variable and its corresponding probability:
E[X] = X
x
x·P(X=x).
Step 2: Given the pmf for random variable X, we can calculate the expected
value as follows:
E[X] = (−3)(0.1) + (0)(0.2) + (1)(0.4) + (2)(0.3).
Step 3: Simplifying the expression yields:
E[X] = −0.3+0+0.4+0.6.
Step 4: Therefore, the expected value of the random variable Xis:
E[X]=0.7.
Question 6
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2 3 4
P(X=x)1
8
1
4
1
201
8
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:
4
E(X) = 0 ·1
8+ 1 ·1
4+ 2 ·1
2+ 3 ·0+4·1
8
Step 3: Simplify the expression:
E(X) = 0 + 1
4+1+0+1
2
E(X) = 1
4+2
2+1
2
Step 4: Combine the fractions:
E(X) = 1
4+2
2+1
2=1
4+1+1
2=1
4+2
4+2
4=5
4
Step 5: Therefore, the expected value of Xis 5
4.
Question 7
Question
Let Xbe a discrete random variable with probability mass function given by
P(X=−2) = 0.1, P (X= 0) = 0.4, P (X= 1) = 0.3, P (X= 3) = 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
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 and calculate the
expected value.
E(X) = (−2) ·0.1 + (0) ·0.4 + (1) ·0.3 + (3) ·0.2
E(X) = −0.2+0+0.3+0.6
E(X) = 0.7.
Therefore, the expected value of the random variable Xis 0.7.
5
Question 8
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
8, P (X= 0) = 1
4, P (X= 1) = 1
2, P (X= 2) = 1
8
Find 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 given values into the formula to find E(X):
E(X)=(−2) ·1
8+ (0) ·1
4+ (1) ·1
2+ (2) ·1
8
Step 3: Simplify the expression:
E(X) = −2
8+0+1
2+2
8=−2
8+4
8+1
2+2
8=5
8+4
8=9
8
Step 4: Therefore, the expected value of Xis 9
8.
Question 9
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
8, P (X= 0) = 2
8, P (X= 3) = 5
8
Find the expected value of X.
6
Solution
Step 1: Recall that the expected value of a discrete random variable Xis given
by:
E(X) = X
i
xi·P(X=xi)
Step 2: We will calculate E(X) using the given probability mass function.
Step 3: Substitute the values of xiand P(X=xi) into the formula:
E(X) = (−2) ·1
8+ 0 ·2
8+ 3 ·5
8
Step 4: Simplify the expression:
E(X) = −2
8+0+15
8=13
8
Step 5: Therefore, the expected value of the random variable Xis 13
8.
Question 10
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=x) =
0.1 for x= 1
0.2 for x= 2
0.3 for x= 3
0.4 for x= 4
0 otherwise
Find the expected value E[X] of X.
Solution
Step 1: Recall that the expected value of a discrete random variable Xis cal-
culated as E[X] = Px·P(X=x) over all possible values of X.
Step 2: In this case, we have the PMF P(X=x) defined for x= 1,2,3,4.
Step 3: We can now calculate the expected value E[X] by multiplying each
possible value of Xby its corresponding probability and summing the results:
E[X] = X
x
x·P(X=x)=1·0.1+2·0.2+3·0.3+4·0.4
Step 4: Now, we can compute the expected value:
E[X]=0.1+0.4+0.9+1.6=3
Therefore, the expected value of the random variable Xis E[X] = 3.
7
Question 11
Question
Let Xbe a discrete random variable with the following probability distribution:
x012
P(X=x) 0.2 0.5 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
i
xi·P(X=xi)
where xiare the possible values of Xand P(X=xi) are their respective
probabilities.
Step 2: Calculate the expected value E(X) using the provided probability
distribution:
E(X)=0·0.2+1·0.5+2·0.3
Step 3: Simplify the expression:
E(X) = 0 + 0.5+0.6
E(X)=1.1
Step 4: Therefore, the expected value of the random variable Xis E(X) =
1.1.
Question 12
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 1
9, P (X= 0) = 2
9, P (X= 3) = 2
9, P (X= 5) = 4
9
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) = Xx·P(X=x)
Step 2: Calculate the expected value E(X) by summing the products of each
value of Xwith its corresponding probability:
E(X)=(−2) ·1
9+ 0 ·2
9+ 3 ·2
9+ 5 ·4
9
Step 3: Simplify the expression:
E(X) = −2
9+0+6
9+20
9=24
9
Step 4: Reduce the fraction:
E(X) = 8
3
Therefore, the expected value of the random variable Xis 8
3.
Question 13
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) = 0.4, P (X= 3) = 0.2.
Calculate the expected value of X.
Solution
Step 1: The expected value, denoted by µ, of a discrete random variable Xis
calculated as:
E[X] = X
x
x·P(X=x),
where the sum is taken over all possible values xthat Xcan take.
Step 2: Substituting the values from the probability mass function into the
formula:
E[X] = (−2) ·0.1 + (0) ·0.3 + (1) ·0.4 + (3) ·0.2
9
Step 3: Calculating the expected value:
E[X] = −0.2+0+0.4+0.6 = 0.8.
Therefore, the expected value of the random variable Xis 0.8.
Question 14
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= 2) = 0.3, P (X= 4) = 0.2
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 of X.
Step 2: Substitute the given probabilities into the formula:
E(X)=(−2)(0.1) + (0)(0.4) + (2)(0.3) + (4)(0.2)
Step 3: Perform the calculations:
E(X) = −0.2+0+0.6+0.8 = 1.2
Step 4: Therefore, the expected value of the random variable Xis 1.2 .
Question 15
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2
P(X=x) 0.3 0.4 0.3
Calculate the expected value of X.
10
Solution
The expected value of a discrete random variable Xis given by:
E(X) = X
all x
x·P(X=x)
Step 1: Calculate the expected value of Xusing the probability distribution
provided.
E(X)=0·0.3+1·0.4+2·0.3
= 0 + 0.4+0.6
= 1
Step 2: Therefore, the expected value of the random variable Xis 1.
Question 16
Question
Let Xbe a discrete random variable with the following probability distribution:
X−2 0 1 3
P(X) 0.2 0.3a0.1
If E[X] = 0.6, find the value of a.
Solution
Step 1: Recall that the expected value E[X] of a discrete random variable Xis
given by the formula:
E[X] = X
x
x·P(X=x)
Step 2: Substitute the given values into the formula for E[X] and solve for
E[X]:
E[X]=(−2)(0.2) + (0)(0.3) + (1)(a) + (3)(0.1)
0.6 = −0.4 + 0 + a+ 0.3
Step 3: Simplify the equation:
0.6 = −0.1 + a
a= 0.6+0.1=0.7
Step 4: Therefore, the value of ais 0.7.
11
Question 17
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
0.4 if x= 4
0 otherwise
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: Calculate the expected value of Xusing the probability mass func-
tion provided:
E(X) = (1)(0.2) + (2)(0.3) + (3)(0.1) + (4)(0.4)
Step 3: Simplify the expression by multiplying the values of xby their
respective probabilities:
E(X)=0.2+0.6+0.3+1.6
Step 4: Add up the terms to find the expected value of X:
E(X)=2.7
Therefore, the expected value of the random variable Xis 2.7.
Question 18
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 1
5, P (X= 0) = 2
5, P (X= 1) = 2
5
Calculate the expected value of X.
12
Solution
Step 1: First, 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: Using the probability mass function provided, we can calculate the
expected value as follows:
E(X) = −1·1
5+ 0 ·2
5+ 1 ·2
5
Step 3: Simplifying the expression, we get:
E(X) = −1
5+0+2
5=1
5
Step 4: Therefore, the expected value of the random variable Xis 1
5.
Question 19
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) = 0.2, P (X= 2) = 0.4
Find the expected value of X.
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
as E(X), is calculated as:
E(X) = Xx·P(X=x)
Step 2: Substituting the given probability mass function values, we have:
E(X) = (−1) ·0.1 + (0) ·0.3 + (1) ·0.2 + (2) ·0.4
Step 3: Calculating the expected value:
E(X)=(−0.1) + (0) + (0.2) + (0.8)
E(X)=0.9
Therefore, the expected value of Xis 0.9.
13
Question 20
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.
Calculate the expected value of X.
Solution
To find the expected value of a discrete random variable X, denoted as E(X),
we use the formula:
E(X) = X
all x
x·P(X=x).
Step 1: Calculate the expected value E(X).
E(X)=1·0.2+2·0.3+3·0.4+4·0.1
Step 2: Simplify the expression.
E(X)=0.2+0.6+1.2+0.4
E(X)=2.4
Therefore, the expected value of the random variable Xis 2.4.
Question 21
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=k) =
0.2 if k= 1
0.3 if k= 2
0.1 if k= 3
0.4 if k= 4
Find the expected value of X.
14
Solution
1. First, recall that the expected value of a discrete random variable Xis
given by the formula:
E(X) = X
k
k·P(X=k)
where the sum is taken over all possible values of X.
2. Now, we can calculate the expected value of Xusing the given probability
mass function:
E(X) = X
k
k·P(X=k) = (1)(0.2) + (2)(0.3) + (3)(0.1) + (4)(0.4)
3. Now, we can plug in the values and simplify:
E(X) = 1(0.2) + 2(0.3) + 3(0.1) + 4(0.4) = 0.2+0.6+0.3+1.6 = 2.7
4. Therefore, the expected value of the random variable Xis 2.7.
Question 22
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.3, P (X= 0) = 0.4, P (X= 3) = 0.1, P (X=k)=0.2
Find the expected value of X.
Solution
Step 1: First, 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: We can now calculate the expected value of Xusing the probability
mass function provided.
E[X] = (−1)(0.3) + (0)(0.4) + (3)(0.1) + (k)(0.2)
Step 3: Simplifying, we get:
E[X] = −0.3+0+0.3+0.2k
15
Step 4: Further simplifying, we find:
E[X]=0.2k
Therefore, the expected value of the discrete random variable Xis 0.2k.
Question 23
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: The expected value of a discrete random variable Xis defined as:
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 for expected value:
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=3
Therefore, the expected value of the random variable Xis 3.
Question 24
Question
Let Xbe a discrete random variable with the following probability distribution:
X1 2 3 4
P(X) 0.1 0.3 0.4 0.2
Find the expected value of X.
16
Solution
Step 1: Recall that the expected value of a discrete random variable X, denoted
by E(X), is given by 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: Calculate the expected value of Xusing the probability distribution
given:
E(X)=1·0.1+2·0.3+3·0.4+4·0.2
Step 3: Perform the calculations to find the expected value:
E(X) = 0.1+0.6+1.2+0.8=2.7
Step 4: Therefore, the expected value of the random variable Xis 2.7 .
Question 25
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
Calculate the expected value of X.
Solution
To find the expected value of a discrete random variable X, we use the formula:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 1: Calculate the expected value
The expected value E(X) of Xis given by:
E(X) = 1 ·P(X= 1) + 2 ·P(X= 2) + 3 ·P(X= 3)
Substitute the given probabilities:
E(X) = 1 ·1
4+ 2 ·1
4+ 3 ·1
2
E(X) = 1
4+1
2+3
2
17
E(X) = 1
4+2
4+6
4
E(X) = 9
4
Therefore, the expected value of the random variable Xis 9
4.
Question 26
Question
Let Xbe a discrete random variable with the following probability distribution:
x P (X=x)x2P(X=x)
0 0.2 0
1 0.3 0.3
2 0.4 0.8
3 0.1 0.3
Calculate the expected value of X.
Solution
To find the expected value of X, denoted by E(X), we use the formula:
E(X) = X
x
x·P(X=x)
Given the probability distribution of X, we have:
E(X) = X
x
x·P(X=x)=0·0.2+1·0.3+2·0.4+3·0.1
Step 1: Calculate E(X)
E(X) = 0 ·0.2+1·0.3+2·0.4+3·0.1 = 0 + 0.3+0.8+0.3=1.4
Therefore, the expected value of Xis 1.4.
Question 27
Question
Let Xbe a discrete random variable with the following probability distribution:
x0 1 2
P(X=x) 0.2 0.5p
If E(X)=1.4, find the value of p.
18
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: In this case, we are given that E(X) = 1.4, so we can write the
expression for the expected value as:
E(X) = 0 ·0.2+1·0.5+2·p
Step 3: Now, we substitute E(X)=1.4 into the above equation:
1.4 = 0 + 0.5+2p
Step 4: Simplifying the equation, we get:
1.4 = 0.5+2p
Step 5: Subtracting 0.5 from both sides gives:
0.9=2p
Step 6: Finally, solving for pgives:
p=0.9
2= 0.45
Therefore, the value of pis 0.45.
Question 28
Question
Let Xbe a discrete random variable with the following probability distribution:
x−1 0 3
P(X=x)1
4
1
2
1
4
Find 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: Substitute the values of xand P(X=x) into the formula. We have:
19
E(X)=(−1) ·1
4+ 0 ·1
2+ 3 ·1
4
Step 3: Simplify the expression:
E(X) = −1
4+0+3
4
E(X) = 2
4
Step 4: Further simplify the fraction:
E(X) = 1
2
Therefore, the expected value of the random variable Xis 1
2.
Question 29
Question
Let Xbe a discrete random variable with the following probability distribution:
X0 1 2
P(X)1
3
1
2
1
6
Calculate the expected value of X, denoted as E[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 that Xcan take and P(X=xi) is the
probability of Xtaking the value xi.
Step 2: In this case, we have:
E[X]=0·1
3+ 1 ·1
2+ 2 ·1
6
Step 3: Simplify the expression:
E[X] = 0 + 1
2+1
3
Step 4: Calculate the sum:
E[X] = 1
2+1
3
20
Step 5: Find a common denominator:
E[X] = 3
6+2
6
Step 6: Add the fractions:
E[X] = 5
6
Therefore, the expected value of the random variable Xis E[X] = 5
6.
Question 30
Question
Let Xbe a discrete random variable with the following probability distribution:
x−2 0 1
P(X=x) 0.3 0.4 0.3
Find 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: Calculate the expected value of Xusing the given probability dis-
tribution:
E(X)=(−2) ·0.3 + (0) ·0.4 + (1) ·0.3
Step 3: Simplify the expression:
E(X) = −0.6+0+0.3
Step 4: Calculate the final answer:
E(X) = −0.3
Therefore, the expected value of the random variable Xis −0.3 .
21
Question 31
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−1) = 0.1, P (X= 0) = 0.2, P (X= 1) = 0.3, 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 given probability mass function into the formula for
expected value:
E(X)=(−1)(0.1) + (0)(0.2) + (1)(0.3) + (2)(0.4)
Step 3: Calculate the expected value:
E(X) = −0.1+0+0.3+0.8
E(X) = 1
Therefore, the expected value of the random variable Xis 1.
Question 32
Question
Let Xbe a discrete random variable with the following probability mass func-
tion:
P(X=−2) = 0.1, P (X= 0) = 0.3, P (X= 3) = 0.6
Calculate the expected value of X.
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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: Using the given probability mass function for X, we can calculate
the expected value as follows:
E(X)=(−2) ·0.1 + (0) ·0.3 + (3) ·0.6
Step 3: Simplifying, we get:
E(X) = −0.2+0+1.8
Step 4: Therefore, the expected value of Xis:
E(X)=1.6
Hence, the expected value of the random variable Xis 1.6.
Question 33
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
Find the expected value of X.
Solution
Step 1: The expected value of a discrete random variable X, denoted by E(X),
is defined by the formula:
E(X) = Xx·P(X=x)
Step 2: Given the probability mass function, we can calculate the expected
value of Xas follows:
E(X) = 1 ×0.1+2×0.2+3×0.3+4×0.4
Step 3: Simplifying the expression, we get:
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E(X)=0.1+0.4+0.9+1.6
E(X) = 3
Step 4: Therefore, the expected value of the random variable Xis 3 .
Question 34
Question
Let Xbe a discrete random variable with the probability mass function given
by:
P(X=−1) = 1
4, P (X= 0) = 1
2, P (X= 2) = 1
4
Calculate 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 defined as:
E(X) = X
x
x·P(X=x)
where the sum is taken over all possible values of X.
Step 2: In this case, the possible values of Xare -1, 0, and 2, with their
corresponding probabilities as given. We will substitute these values into the
formula for expected value and calculate the result:
E(X)=(−1) ·1
4+ (0) ·1
2+ (2) ·1
4
Step 3: Simplifying the expression, we get:
E(X) = −1
4+0+1
2
E(X) = 1
4
Therefore, the expected value of the random variable Xis 1
4.
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Question 35
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= 2) = 0.4.
Calculate the expected value of X.
Solution
Step 1: To calculate the expected value of a discrete random variable X, we use
the formula:
E(X) = X
all x
xP (X=x).
Step 2: We substitute the given probabilities for each value of Xinto the
formula:
E(X)=(−2)(0.1) + (0)(0.5) + (2)(0.4).
Step 3: Calculate the expected value:
E(X) = −0.2+0+0.8 = 0.6.
Therefore, the expected value of the random variable Xis 0.6.
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