Binomial Probability Problems

Binomial Probability Problems
• Page 1
1.
What is the probability that a month selected in a year starts with letter 'J'?
 a. $\frac{1}{4}$ b. 1 c. $\frac{1}{6}$ d. zero

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 = { January, February, March, April, May, June, July, August, September, October, November, December }.

Number of elements in sample space n(S) = 12

Months that start with a letter 'J' = { January, June, July }.

Number of months that start with a letter 'J' n(E) = 3

P(E) = 3 / 12
[Substitute the values of n(E) and n(S) in P(E).]

P(E) = 1 / 4
[Simplify.]

The probability that a month selected in a year will start with letter 'J' is 1 / 4 .

2.
What is the probability that a day selected in a week starts with letter 'S'?
 a. $\frac{3}{7}$ b. one c. $\frac{2}{7}$ d. $\frac{1}{7}$

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 = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}.

Number of elements in sample space n(S) = 7.

Days of week that start with a letter 'S' = {Sunday, Saturday}

P(E) = 2 / 7
[Substitute the values of n(E) and n(S) in P(E).]

The probability that a day selected in a week will start with letter 'S' is 2 / 7.

3.
If 2 coins are tossed, then what is the probability of getting at least 1 head?
 a. $\frac{1}{4}$ b. one c. $\frac{1}{2}$ d. $\frac{3}{4}$

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)

Let us denote the occurence of Head with 'H' and Tail with 'T'.

Sample space = { HH, HT, TH, TT }.

Number of elements in the sample space n(S) = 4

Number of times a head occurs at least once n(E) = 3

P(E) = 3 / 4
[Substitute the values of n(E) and n(S) in P(E).]

The probability of getting at least 1 head when 2 coins are tossed is 3 / 4 .

4.
A probability experiment is conducted. Which of these cannot be considered a probability of an outcome?
(i) 0.7
(ii) 0
(iii) 1.2
(iv) - 1.6
 a. (iv) only b. (ii) only c. (iii) only d. (iii) and (iv)

Solution:

The probability of any event E is a number between and including 0 and 1.
Probability cannot be negative or greater than 1.

Consider the given choices.
0.7 lies between 0 and 1. It is a possible outcome.

0 satisfies the condtion. It is a possible outcome.

1.2 is greater than 1. It is not a possible outcome.

-1.6 is negative. It is not a possible outcome.

So (iii) and (iv) cannot be considered a probability of an outcome.

5.
If 2 coins are tossed, then what is the probability of not getting any head?
 a. $\frac{3}{4}$ b. $\frac{1}{4}$ c. $\frac{1}{2}$ 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)

Let us denote the occurence of Head with 'H' and Tail with 'T'.

Sample space = { HH, HT, TH, TT }.

Number of elements in the sample space n(S) = 4.

Number of times a head does not occur at least once n(E) = 1

P(E) = 1 / 4
[Substitute the values of n(E) and n(S) in P(E).]

The probability of not getting any head when 2 coins are tossed is 1 / 4 .

6.
When a single die is rolled, what is the probability of getting a 9?
 a. $\frac{1}{6}$ b. zero c. $\frac{1}{2}$ 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)

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

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

We do not get a 9 when a single die is rolled, so n(E) = 0

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

The probability of getting a 9 when a single die is rolled is 0.

7.
When a single die is rolled, what is the probability of getting a number less than 7?
 a. $\frac{1}{2}$ b. one c. zero d. $\frac{1}{6}$

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

We find that all the elements of the sample space are less than 7, so n(E) = 6.

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

P(E) = 1
[Simplify.]

The probability of getting a number less than 7 when a single die is rolled is 1.

8.
When a single die is rolled, what is the probability of getting an even number?
 a. $\frac{1}{2}$ b. One c. $\frac{1}{3}$ d. $\frac{2}{3}$

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 numbers in the sample space = { 2, 4, 6 }.

Number of even numbers n(E) = 3

P(E) = 3 / 6
[Substitute the values of n(E) and n(S) in P(E).]

P(E) = 1 / 2
[Simplify.]

The probability of getting an even number when a single die is rolled is 1 / 2 .

9.
What is the probability of getting an odd number when a single die is rolled?
 a. $\frac{1}{2}$ b. one c. $\frac{1}{3}$ d. $\frac{2}{3}$

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

Odd numbers in the sample space = { 1, 3, 5 }.

Number of odd numbers n(E) = 3

P(E) = 3 / 6
[Substitute the values of n(E) and n(S) in P(E).]

P(E) = 1 / 2
[Simplify.]

The probability of getting a odd number when a single die is rolled is 1 / 2.

10.
When a single die is rolled, what is the probability of getting a prime number?
 a. one b. $\frac{2}{3}$ c. $\frac{1}{2}$ d. $\frac{1}{3}$

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

Prime numbers in the sample space = { 2, 3, 5 }.

Number of prime numbers n(E) = 3

P(E) = 3 / 6
[Substitute the values of n(E) and n(S) in P(E).]

P(E) = 1 / 2
[Simplify.]

Probability of getting a prime number when a single die is rolled is 1 / 2 .