$\begin{array}{1 1}(A)\;\large\frac{40}{143}\\(B)\;\large\frac{20}{143}\\(C)\;\large\frac{70}{143}\\(D)\;\large\frac{10}{143}\end{array} $

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Step 1:

Given a bag contains 8 red and 5 white balls

$\therefore$ The sample space is $S=13(8+5)$

Step 2:

One ball is red and two balls are white

Three cases arise in this situation

Case I:

If red balls is drawn at first position and then the white balls

$P=\large\frac{8}{13}\times \frac{5}{12}\times \frac{4}{11}$

$\;\;\;=\large\frac{40}{3\times 143}$

Step 3:

Case II:

If red ball is drawn at second position and then the white balls correspondingly.

$P=\large\frac{5}{13}\times \frac{8}{12}\times \frac{4}{11}$

$\;\;\;=\large\frac{40}{3\times 143}$

Step 4:

Case III:

If red ball is drawn at third position

$P=\large\frac{5}{13}\times \frac{4}{12}\times \frac{8}{11}$

$\;\;=\large\frac{40}{3\times 143}$

Step 5:

P[one ball is red and two balls are white]

$\Rightarrow \large\frac{40}{3\times 143}+\frac{40}{3\times 143}+\frac{40}{3\times 143}$

$\Rightarrow \large\frac{3\times 40}{3\times 143}$

$\Rightarrow \large\frac{ 40}{143}$

Hence (A) is the correct answer.

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