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. 14/55
• B. 12/55
• C. 2/11
• D. 8/55

Answer 1

The correct option is A

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
$P\left(E_i\right)=\frac{1}{11}(i=0,1,2,....,10)P\left(A/E_i\right)=0\text{for}i=0,1,2\text{and}P\left(A/E_i\right)=\frac{i_{C_3}}{{10}_{C_3}}\text{for}i\ge 3$
Now, by the total probability rule
$P\left(A\right)={\sum }_{i=0}^{10}P\left(E_i\right)P\left(A/E_i\right)=\frac{1}{11}\times \frac{1}{{10}_{C_3}}[3_{C_3}+4_{C_3}+....+{10}_{C_3}]But{3}_{C_3}+4_{C_3}+....+{10}_{C_3}==4_{C_4}+4_{C_3}+....+{10}_{C_3}=5_{C_4}+5_{C_3}+....+{10}_{C_3}=6_{C_4}+6_{C_3}+....+{10}_{C_3}=....={11}_{C_4}$
Thus, $P\left(A\right)=\frac{{11}_{C_4}}{11\times {10}_{C_3}}=\frac{1}{4}$
By the Baye's rule
$P\left(E_9/A\right)=\frac{P\left(E_9\right)P\left(A/E_9\right)}{P\left(A\right)}=\frac{\frac{1}{11}\times \frac{9_{C_3}}{{10}_{C_3}}}{\frac{1}{4}}=\frac{14}{55}$