Question

A box contains tickets numbered from 1 to 24. 3 tickets are to be chosen to give 3 prizes. What is the probability that at least 2 tickets contain a number which is multiple of 3 ?

Answer 1

Total possible outcome = 24

We want no. divisible by 3. There are 3,6,9,12,15,18,21,24 i. e. 8 numbers.

P ( getting at least two no. divisible by 3) = 8/24 * 2

= 1/3 * 2

=2/3 or 4/6.

This is the probability.

Answer 2

Answer:

$\frac{63}{253}$

Step-by-step explanation:

From 1 to 24, there are 8 numbers that are multiples of 3.

(24-8) = 16 numbers are not multiples of 3.

Case 1: Two tickets are multiples of 3 and one is not.

$\frac{8C_{2} * 16C_{1} }{24C_{3} }$ = $\frac{56}{253}$

Case 2: All are multiples of 3.

$\frac{8C_{3} }{24C_{3} }$ = $\frac{7}{253}$

Probability that at least 2 tickets contain a number which is multiple of 3 is

$\frac{56}{253} + \frac{7}{253}$

= $\frac{63}{253}$