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# A bag contains 8 red and 5 white balls.Three balls are drawn at random.Find the probability that one ball is red and two balls are white

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

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.