Question

P) A box contains 8 black and 4white balls .if 5 balls are drawn
at random find the Probability that 3 of them
are black & 2 white
​

Answer 1

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

➢ Given that bag contains

• Number of black balls = 8

• Number of white balls = 4

So,

• Total number of balls in bag = 12.

Now,

➢ Number of ways in which 5 balls can be taken out randomly from 12 balls is

$\rm \: \: = \: ^{12}C_5$

$\rm \: \: = \:\dfrac{12!}{5! \: (12 - 5)!}$

$\rm \: \: = \:\dfrac{12!}{5! \: 7!}$

$\rm \: \: = \:\dfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{5 \times 4 \times 3 \times 2 \times 1 \times \: 7!}$

$\rm \: \: = \:792$

Now,

➢ Number of ways in which 3 black balls can be drawn from 8 black balls is

$\rm \: \: = \: ^{8}C_3$

$\rm \: \: = \:\dfrac{8!}{3! \: (8 - 3)!}$

$\rm \: \: = \:\dfrac{8!}{3! \: 5!}$

$\rm \: \: = \:\dfrac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times \: 5!}$

$\rm \: \: = \:56$

Again,

➢ Number of ways in which 2 white balls can be drawn randomly from 4 white balls is

$\rm \: \: = \: ^{4}C_2$

$\rm \: \: = \:\dfrac{4!}{2! \: (4 - 2)!}$

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

$\rm \: \: = \:\dfrac{12}{ 2\times 1}$

$\rm \: \: = \:6$

So,

➢ Required probability of getting 3 black balls and 2 white balls is given by

$\rm \: \: = \:\dfrac{ ^8C_3 \: \times \: ^4C_2 }{ ^{12}C_5 }$

$\rm \: \: = \:\dfrac{6 \times 56}{792}$

$\rm \: \: = \:\dfrac{56}{132}$

$\rm \: \: = \:\dfrac{14}{33}$

Formula Used :-

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

Additional Information :-

$\rm :\longmapsto\:P(A\cup \:B) = P(A) + P(B) - P(A\cap \:B)$

$\rm :\longmapsto\:P(A'\cap \:B) = P(B) - P(A\cap \:B)$

$\rm :\longmapsto\:P(A\cap \:B') = P(A) - P(A\cap \:B)$

$\rm :\longmapsto\:P(A'\cap \:B') = 1 - P(A\cup \:B)$

$\rm :\longmapsto\:P(A'\cup \:B') = 1 - P(A\cap \:B)$