Question

A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is

• A. $\frac{8}{55}$
• B. $\frac{2}{55}$
• C. $\frac{12}{55}$
• D. $\frac{14}{55}$

Answer 1

Answer: (D)

Let $E_i$ denotes the event when bag contain $i$ black balls and $(10-i)$ white balls i.e., $i\in \{0,1,2,\dots \dots ,10\}$
Total 11 events are possible As, each event/combination is equally likely
$\Rightarrow P\left(E_i\right)=\frac{1}{11}$
Let Adenotes the event of selecting 3 Black Balls.
$\therefore P\left(\frac{A}{E_i}\right)=\frac{1}{11}\times \frac{{}^i{C}_3}{{}^{10}{C}_3}$
$\text{ Total probability of selecting }3\text{ black balls }=\sum P\left(\frac{A}{E_i}\right)$
$=\frac{1}{11}\left(0+0+0+\frac{{}^3{C}_3}{{}^{10}{C}_3}+\frac{{}^4{C}_3}{{}^{10}{C}_3}+\dots ..+\frac{{}^{10}{C}_3}{{}^{10}{C}_3}\right)=\frac{1}{11}\left(\frac{{}^3{C}_3+{}^4{C}_3+\dots ..{}^{10}{C}_3}{{}^{10}{C}_3}\right)$
$=\frac{1}{11}\left(\frac{{}^4{C}_4+{}^4{C}_3+\dots \dots ..{}^{10}{C}_3}{{}^{10}{C}_3}\right)=\frac{1}{11}\times \frac{{}^{11}{C}_4}{{}^{10}{C}_3}$
Required probability $=P\left(\frac{E_9}{A}\right)=\frac{P\left(A\cap E_9\right)}{\sum P\left(\frac{A}{E_i}\right)}=\frac{\frac{1}{11}\times \frac{{}^9{C}_3}{{}^{10}{C}_3}}{\frac{1}{11}\times \frac{{}^{11}{C}_4}{{}^{10}{C}_3}}=\frac{\frac{9\times 8\times 7}{6}}{\frac{11\times 10\times 9\times 8}{24}}=\frac{14}{55}$