Question

You have been given 60 balls, which are labelled as 1,2,3...,60 respectively

a. What is the probability of pulling out a number that is a multiple of 7 provided the pulled number is an even one ?

b. What is the probability of drawing a number that is a multiple of 4 given the fact that the drawn number is a multiple of 3?

Answer 1

(a)

Let

A = multiple of 7

Favorable events = 7, 14, 21, 28, 35, 42, 49, 56: 8 Nos.

So,

P(A) = 8/60 = 0.1333

B = even number

So,

P(B) = 30/60 = 0.5

AB = multiple of 7 AND even number:

Favorable events = 14, 28, 42, 56: 4 Nos.

So,

P(AB) = 4/60 = 0.0667

By Multiplication Theorem:
P(A/B) = P(AB)/P(B)

            = 0.0667/0.5 = 0.1333

So,

Answer is:

0.1333

(b)

Let

A = multiple of 4

Favorable events = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60: 15 Nos.

So,

P(A) = 15/60 = 0.25

B = multiple of 3

Possible events: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60: 20 Nos.

So,

P(B) = 20/60 = 0.3333

AB = multiple of 4 AND multiple of 3

Favorable events = 12, 24, 36, 48, 60: 5 Nos.

So,

P(AB) = 5/60 = 0.0833

By Multiplication Theorem:
P(A/B) = P(AB)/P(B)

            = 0.0833/0.3333 = 0.25

So,

Answer is:

0.25