Question

A bag contains 4 red marbles, 5 green marbles and 6 pink marbles. If 3 marbles are taken at randomly, then find the probability that 2 marbles are Pink? a) 27/91 b) 45/77 c) 34/105 d) 11/23 e) None of these​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A bag contains 4 red marbles, 5 green marbles and 6 pink marbles.

Total number of marbles = 4 + 5 + 6 = 15 marbles.

It is further given that, 3 marbles are taken out randomly.

So, number of ways in which 3 marbles can be drawn from 15 marbles =

$\rm \: = \: ^{15}C_{3}$

$\rm \: = \: \dfrac{15!}{3! \: (15 - 3)!}$

$\rm \: = \: \dfrac{15!}{3! \:12!}$

$\rm \: = \: \dfrac{15 \times 14 \times 13 \times 12!}{3 \times 2 \times \:12!}$

$\rm \: = \: 35 \times 13$

$\rm \: = \: 455$

Now, we have to find the probability of getting 2 pink marbles when 3 marbles are taken out randomly.

It means, we have to find the probability of getting 2 pink marbles and one non pink marbles when 3 marbles are taken out randomly

So, Number of ways in which 2 pink marbles can be drawn from 6 pink marbles and 1 marble from remaining 9 marbles is

$\rm \: = \: ^{6}C_{2} \: \times \: ^{9}C_{1}$

$\rm \: = \: \dfrac{6!}{2! \: (6 - 2)!} \times \dfrac{9!}{1! \: (9 - 1)!}$

$\rm \: = \: \dfrac{6!}{2! \: 4!} \times \dfrac{9!}{8!}$

$\rm \: = \: \dfrac{6 \times 5 \times 4!}{2 \times \: 4!} \times \dfrac{9 \times 8!}{8!}$

$\rm \: = \: 15 \times 9$

$\rm \: = \: 135$

So, Required probability of getting 2 pink marbles when 3 marbles are taken out randomly is

$\rm \: = \: \dfrac{135}{455}$

$\rm \: = \: \dfrac{27}{91}$

Hence, Option (a) is correct.

$\rule{190pt}{2pt}$

Formula Used :-

$\boxed{ \rm{ \:^{n}C_{r} \: = \: \frac{n!}{r! \: (n - r)!} \: \: }}$