Question

A bag contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that none is white all are white at least one is white only 2 are white

Answer 1

$\underline{\underline{\Huge{\textbf{Solution:}}}}$
Given,
A bag contains 5 White, 7 Red , 8 Black balls.

$\textbf{Experiment}$- 4 balls are drawn One by one with replacement.

$\textbf{Events:}$

Let A be the Event that None of the ball drawn is red.
Let B be the Event that all the balls drawn are white.
Let C be the Event that balls drawn contains at least one white
Let D be the Event that the balls drawn contains 2 White balls.

$\begin{array}{cccc}P(A) = ?& P(B)= ?& P(C) = ? &P(D)=?\end{array}$



$\underline{\LARGE{\textsf{Step-1:}}}$
$\textsf{Note the total No. of balls in bag.}$

Total No. of balls in bag = No. of White balls+ No. of Red balls+No. of Black balls

Total No. of balls in bag = 5+7+8= 20.

As the Experiment is done $\textit{with replacement}$, the $\textsf{total number of balls remained}$ after drawing each ball from the bag will be $\textsf{20}$.
($\because$ The drawn ball is replaced in the bag)

$\underline{\LARGE{\textsf{Step-2:}}}$
$\textsf{Note the total No. of Outcomes in Experiment.}$

Total No. of Outcomes = No. of ways of drawing 4 balls from the bag with replacement

Total No. of Outcomes = $20 \times 20\times 20\times 20$

$\therefore \textbf{Total No. of Outcomes n(S) = 20^4 }$

$\underline{\LARGE{\textsf{Step-3:}}}$
$\textsf{Find Favorable Outcomes for Occurrence}\\\textsf{of Event A.}$

If the balls drawn must not contain White balls, the balls drawn may contain Non-White balls.

No. of ways of drawing a Non-White ball = No. of red balls + No. of Black balls

Favorable Outcomes for drawing a non white ball = 7+8 = 15

$\therefore \textbf{Favorable Outcomes for Event A, n(A) = 15^4 }$
($\because$ 4 balls are drawn)

$\underline{\LARGE{\textsf{Step-4:}}}$
$\textsf{Find Favorable Outcomes for Occurrence}\\\textsf{of Event B.}$

Favorable Outcomes for drawing a white ball =No. of white balls.

Favorable Outcomes for drawing a white ball = 5

$\therefore \textbf{Favorable Outcomes for Event B, n(B) = 5^4 }$
($\because$ 4 balls are drawn)

$\underline{\LARGE{\textsf{Step-5:}}}$
$\textsf{Find Favorable Outcomes for Occurrence}\\\textsf{of Event D.}$

If the balls drawn must contain only two White balls, the remaining balls drawn should contain Non-white balls.

Favourable outcomes for Event D = No. of ways of drawing 2 White balls $\times$ No. of ways of drawing 2 Non-White balls.

$\therefore \textbf{Favourable outcomes for Event D, n(D)= 5^2 \times 15^2 .}$



We have,

$\bigstar\; \boxed{\mathbf{Probability\: of\: Occurrence\: of\: Event = \dfrac{Favorable\: Outcomes\: for\: Event}{Total\: No.\: of\: outcomes\: in\: Experiment}}}$



$\underline{\LARGE{\textsf{Step-6:}}}$
$\textsf{Find Probability of occurrence of Event A}$

$\mathbf{P(A) = \dfrac{n(A)}{n(S)}}$

$\implies \mathsf{P(A) = \dfrac{15^4}{20^4}}$

$\implies\mathsf{P(A) = \dfrac{(5\times3)^4}{(5\times4)^4}}$

$\implies\mathsf{P(A) = \dfrac{\cancel{5^4}\times 3^4}{\cancel{5^4}\times 4^4}}$

$\implies\mathsf{P(A) = \dfrac{3^4}{4^4}}$

$\therefore \textsf{Probability that none is white}\; \boxed{\mathbf{P(A) = \dfrac{81}{256}}}$

$\underline{\LARGE{\textsf{Step-7:}}}$
$\textsf{Find Probability of occurrence of Event B}$

$\mathbf{P(B) = \dfrac{n(B)}{n(S)}}$

$\implies \mathsf{P(B) = \dfrac{5^4}{20^4}}$

$\implies \mathsf{P(B) = \dfrac{5^4}{(5\times4)^4}}$

$\implies\mathsf{P(B) = \dfrac{\cancel{5^4}}{\cancel{5^4}\times4^4}}$

$\implies\mathsf{P(B) = \dfrac{1}{4^4}}$

$\therefore\textsf{Probability that all are white}\; \boxed{\mathbf{P(B) = \dfrac{1}{256}}}$

$\underline{\LARGE{\textsf{Step-8:}}}$
$\textsf{Find Probability of occurrence of Event D}$

$\mathbf{P(D) = \dfrac{n(D)}{n(S)}}$

$\implies\mathsf{P(D) = \dfrac{5^2\times15^2}{20^4}}$

$\implies \mathsf{P(D) = \dfrac{5^2\times(5\times3)^2}{(5\times4)^4}}$

$\implies \mathsf{P(D) = \dfrac{5^2\times5^2\times 3^2}{5^4\times4^4}}$

$\implies\mathsf{P(D) = \dfrac{\cancel{5^4}\times 3^2}{\cancel{5^4}\times4^4}}$

$\therefore \textsf{Probability that only two are white}\; \boxed{\mathbf{P(D) = \dfrac{9}{256}}}$

$\underline{\LARGE{\textsf{Step-9:}}}$
$\textsf{Find Probability of occurrence of Event C}$

$\mathbf{P(At least one white ball) = 1 - P(None of ball is white)}$

$\implies \mathsf{P(C) = 1- P( A)}$

$\implies\mathsf{P(C) = 1- \dfrac{81}{256}}$

$\implies P(C) = \dfrac{175}{256}$

$\therefore \textsf{Probability that at least one is white}\; \boxed{\mathbf{P(C) = \dfrac{175}{256}}}$