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Binomial Probability Problems - Page 2

Binomial Probability Problems
  • Page 2
 11.  
When a single die is rolled, what is the probability of getting an even prime number?
a.
1 2
b.
one
c.
1 6
d.
1 3


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Sample space = { 1, 2, 3, 4, 5, 6 }

Number of elements in the sample space n(S) = 6

Even prime number in the given set = { 2 }

Number of even prime numbers n(E) = 1

P(E) = 1 / 6
[Substitute the values of n(E) and n(S) in P(E).]

Probability of getting an even prime number when a single die is rolled is 1 / 6 .


Correct answer : (3)
 12.  
If 2 dice are rolled once, then what is probability of getting a sum of 1?
a.
1 6
b.
one
c.
zero
d.
1 36


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)


Sample space for rolling two dice is

Number of elements in the sample space n(S) = 36

From the table, we see that we can never get a sum of 1. The minimum sum that can be obtained is 2. So n(E) = 0

P(E) = 0.
[Substitute the values of n(E) and n(S) in P(E).]

Probability of getting a sum of 1 when 2 dice are rolled once is 0.


Correct answer : (3)
 13.  
When a single die is rolled, what is the probability of getting an odd number greater than 3?
a.
1 3
b.
1 6
c.
1 2
d.
zero


Solution:

Probabiltiy of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Sample space = { 1, 2, 3, 4, 5, 6 }

Number of elements in the sample space n(S) = 6

Numbers greater than 3 in the sample space = { 4, 5, 6 }

Odd numbers in this set = { 5 }

Number of odd numbers which are greater than 3, n(E) = 1

P(E) = 1 / 6
[Substitute the values of n(E) and n(S) in P(E).]

Probability of getting an odd number greater than 3, when a single die is rolled is 1 / 6 .


Correct answer : (2)
 14.  
If 2 dice are rolled once, then what is probability of getting a sum of 12?
a.
zero
b.
1 3
c.
1 6
d.
1 36


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)


Sample space for rolling two dice is

Number of elements in the sample space n(S) = 36

From the table, we see that to get a sum of 12 both the dice should roll 6. So n(E) = 1

P(E) = 1 / 36
[Substitute the values of n(E) and n(S) in P(E).]

Probability of getting a sum of 12 when 2 dice are rolled once is 1 / 36 .


Correct answer : (4)
 15.  
What is the probability that a selected alphabet is a vowel?
a.
3 13
b.
21 26
c.
5 26
d.
2 13


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Sample space of alphabets = { a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z }

Number of elements in the sample space n(S) = 26

Vowels = { a, e, i, o, u }

Number of vowels n(E) = 5

P(E) = 5 / 26
[Substitute the values of n(E) and n(S) in P(E).]

Probability that a selected alphabet is a vowel is 5 / 26 .


Correct answer : (3)
 16.  
What is the probability that a selected alphabet is a consonant?
a.
19 26
b.
21 26
c.
one
d.
5 26


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Sample space of alphabets = { a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z }

Number of elements in the sample space n(S) = 26

Consonants = { b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z }

Number of consonants n(E) = 21

P(E) = 21 / 26
[Substitute the values of n(E) and n(S) in P(E).]

Probability that a selected alphabet is a consonant is 21 / 26 .


Correct answer : (2)
 17.  
If the probability that it will rain tomorrow is 0.6, then what is the probability that it will not rain tomorrow ?
a.
0.5
b.
0.4
c.
one
d.
0.6


Solution:

The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is E.

P(E) + P(E) = 1

0.6 + P(E) = 1
[Substitute P(E) = 0.6.]

P(E) = 1 - 0.6

P(E) = 0.4

So, the probability that it will not rain tomorrow is 0.4.


Correct answer : (2)
 18.  
A college committee consisted of 4 boys and 3 girls. If a member is selected at random, then what is the probability that the selected member is a boy?
a.
4 7
b.
3 4
c.
3 7
d.
one


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Number of elements in Sample space = Number of boys + Number of girls.

n(S) = 4 + 3

n(S) = 7

Number of boys n(E) = 4

P(E) = 4 / 7
[Substitute the values of n(E) and n(S) in P(E).]

Probability that member selected is a boy is 4 / 7 .


Correct answer : (1)
 19.  
Specify the nature of probability that the following statement denotes:
The probability that the twin children born to a mother are both boys is 1 3.
a.
Cannot be determined
b.
Classical probability
c.
Empirical probability
d.
Subjective probability


Solution:

Classical probability assumes that all outcomes in the sample space are equally likely to occur.

In the given statement outcome has equal probability to occur.

So,the statement denotes classical probability.


Correct answer : (2)
 20.  
A survey was conducted in a class of 50 about the games the students play. Each student had to select only one particular game or no game. The results of the survey are: 20 play football, 15 play basketball and 10 play baseball. What is the probability that a student selected plays football?
a.
3 10
b.
1 10
c.
2 5
d.
3 5


Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Total number of students n(S) = 50

Number of students who play football n(E) = 20

P(E) = 20 / 50
[Substitute the values of n(E) and n(S) in P(E).]

P(E) = 2 / 5
[Simplify.]

Probability that a selected student plays football is 2 / 5 .


Correct answer : (3)

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