Question

what is the probablity of getting 3 white balls in a draw of 3 balls from a box containing 6 white and 5 red balls

Answer 1

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

Given that,

↝ A box having 6 white balls and 5 red balls.

↝ Total number of balls = 6 + 5 = 11 balls

We know,

The number of ways in which r objects can be taken out randomly from n objects is

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_r \: = \: \frac{n!}{r! \: (n - r)!} \: }}}$

So,

Total Number of ways in which 3 balls can be taken out randomly from 11 balls ( 6 white + 5 red balls ) is

$\red{\rm :\longmapsto\:^{11}C_3}$

$\rm \: = \: \dfrac{11!}{3! \: (11 - 3)!}$

$\rm \: = \: \dfrac{11!}{3! \: 8!}$

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

$\rm \: = \: 165$

Now, Number of ways in which 3 white balls can be taken out from 6 white balls is

$\red{\rm :\longmapsto\:^{6}C_3}$

$\rm \: = \: \dfrac{6!}{3! \: (6 - 3)!}$

$\rm \: = \: \dfrac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!}$

$\rm \: = \: 20$

So, Required Probability of getting 3 white balls from 6 white and 5 red balls is

$\rm \: = \: \dfrac{20}{165}$

$\rm \: = \: \dfrac{4}{33}$

Hence,

$\purple{\rm \implies\:\boxed{ \tt{ \: Required \: Probability \: = \: \frac{4}{33} \: }}}$

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Additional Information :-

$\boxed{ \tt{ \: P(A\cup B) = P(A) + P(B) - P(A\cap B) \: }}$

$\boxed{ \tt{ \: P(A\cap B') = P(A) - P(A\cap B) \: }}$

$\boxed{ \tt{ \: P(A'\cap B') = 1 - P(A\cup B) \: }}$

$\boxed{ \tt{ \: P(A'\cup B') = 1 - P(A\cap B) \: }}$

If A and B are independent events then

$\boxed{ \tt{ \: P(A\cap B) = P(A) \: P(B) \: }}$

If A and B are mutually exclusive events, then

$\boxed{ \tt{ \: P(A\cap B) = 0 \: }}$

If A and B are mutually exhaustive events, then

$\boxed{ \tt{ \: P(A\cup B) = 1 \: }}$