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 |

P(E) =

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) =

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

P(E) =

[Simplify.]

The probability that a month selected in a year will start with letter 'J' is

Correct answer : (1)

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. | $\frac{1}{7}$ |

P(E) =

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

Number of elements in sample space

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

Number of days that start with letter 'S'

P(E) =

[Substitute the values of

The probability that a day selected in a week will start with letter 'S' is

Correct answer : (3)

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. | $\frac{3}{4}$ |

P(E) =

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) =

[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

Correct answer : (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

(i) 0.7

(ii) 0

(iii) 1.2

(iv)

a. | (iv) only | ||

b. | (ii) only | ||

c. | (iii) only | ||

d. | (iii) and (iv) |

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.

Correct answer : (4)

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 |

P(E) =

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) =

[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

Correct answer : (2)

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 |

P(E) =

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.

Correct answer : (2)

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. | $\frac{1}{6}$ |

P(E) =

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) =

[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.

Correct answer : (2)

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. | $\frac{2}{3}$ |

P(E) =

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) =

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

P(E) =

[Simplify.]

The probability of getting an even number when a single die is rolled is

Correct answer : (1)

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. | $\frac{2}{3}$ |

P(E) =

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

Number of elements in the sample space

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

Number of odd numbers

P(E) =

[Substitute the values of

P(E) =

[Simplify.]

The probability of getting a odd number when a single die is rolled is

Correct answer : (1)

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. | $\frac{1}{3}$ |

P(E) =

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) =

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

P(E) =

[Simplify.]

Probability of getting a prime number when a single die is rolled is

Correct answer : (3)