Question

Ronnie has a bag containing 7 red and 9 green balls. From it, he draws out 6 balls simultaneously at random. The probability that 4 of them are red and the rest are green is
(A) 24/143
(B) 45/286
(C) 155/2001
(D) 302/1005

Answer 1

Answer:

6/4

Step-by-step explanation:

is the answer of the.

Answer 2

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

Given that,

Ronnie has a bag containing 7 red and 9 green balls.

So, total number of balls in a bag = 9 + 7 = 16

We know,

The number of ways of taking out r objects from n objects is given by

$\boxed{ \bf{ \: ^nC_r = \frac{n!}{r! \: \: (n - r)!}}}$

So,

Number of ways in which 6 balls can be taken out randomly from 16 balls is

$\rm \: = \:^{16}C_6$

$\rm \: = \:\dfrac{16!}{6!(16 - 6)!}$

$\rm \: = \:\dfrac{16!}{6! \: 10!}$

$\rm \: = \:\dfrac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10!}{6 \times 5 \times 4 \times 3 \times 2 \times \: 10!}$

$\rm \: = \:8008$

So, Total number of possible outcomes = 8008

Now,

Number of ways of getting 4 red balls from 7 red balls is

$\rm \: = \:^7C_4$

$\rm \: = \:\dfrac{7!}{4! \: (7 - 4)!}$

$\rm \: = \:\dfrac{7!}{4! \: 3!}$

$\rm \: = \:\dfrac{7 \times 6 \times 5 \times 4!}{4! \times \: 3 \times 2}$

$\rm \: = \:35$

Now,

Number of ways of getting 2 green balls from 9 balls is

$\rm \: = \:^9C_2$

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

$\rm \: = \:\dfrac{9!}{2! \: 7!}$

$\rm \: = \:\dfrac{9 \times 8 \times 7!}{2 \times 1 \times \: 7!}$

$\rm \: = \:36$

So,

Number of favourable outcomes of getting 4 red balls and 2 green balls from 7 red balls and 9 green balls is

$\rm \: = \:36 \times 35$

$\rm \: = \:1260$

So, required Probability is

$\rm \: = \:\dfrac{1260}{8008}$

$\rm \: = \:\dfrac{315}{2002}$