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

Answer 1

The correct option is A $\frac{14}{55}$

Let $E_i$ denote the event that the bag contains $i$ black and $(10-i)$ white balls $(i=0,1,2,...,10)$.

Let $A$ denote the event that the three balls drawn at random from the bag are black. We have
$P\left(E_i\right)=\frac{1}{11}(i=0,1,2,...,10)$
$P\left(A|E_i\right)=0$ for $i=0,1,2$
and $P\left(A|E_i\right)=\frac{{}^i{C}_3}{{}^{10}{C}_3}$ for $i\ge 3$
Now, by the total probability rule
$P\left(A\right)=\frac{1}{11}\times \frac{1}{{}^{10}{C}_3}{[}^3{C}_3{+}^4{C}_3+...{+}^{10}{C}_3]$
But ${}^3{C}_3{+}^4{C}_3{+}^5{C}_3+...{+}^{10}{C}_3$
${=}^4{C}_4{+}^4{C}_3{+}^5{C}_3+...{+}^{10}{C}_3$
${=}^5{C}_4{+}^5{C}_3{+}^6{C}_3+...{+}^{10}{C}_3$
$=....{=}^{11}{C}_4$
Use $P\left(E_9|A\right)=\frac{P\left(E_9\right)P\left(A|E_9\right)}{P\left(A\right)}$$=\frac{1{\times }^9{C}_3}{11{\times }^{10}{C}_3\times \frac{{}^{11}{C}_4}{11{\times }^{10}{C}_3}}$ $=\frac{14}{55}$