Question

Match the following by appropriately matching the lists based on the information given in Column I and Column II.
A bag contains some white and some black balls, all combinations being equally likely. The total number of balls in the bag is 12. Four balls are drawn at random from the bag at random without replacement.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ a.\text{Probability that all the four balls are black is equal to}& p.\frac{14}{33}\\ b.\text{If the bag contains 10 black and 2 white balls, then}& \\ \text{the probability that all four balls are black is equal to}& q.\frac{1}{5}\\ c.\text{If all the four balls are black, then the probability}& \\ \text{that the bag contains 10 black balls is equal to}& r.\frac{70}{429}\end{array}$

• A. $a\to r;b\to q;c\to p;$
• B. $a\to r;b\to q;c\to p;$
• C. $a\to q;b\to p;c\to r;$
• D. $a\to r;b\to p;c\to q;$

Answer 1

The correct option is C $a\to q;b\to p;c\to r;$

Let $E_i$ denote the event that the bag contains $i$ black and $(12-i)$ white balls $(i=0,1,\mathrm{2...},12)$ and $A$ denote the event that the four balls drawn are all black. Then
$P\left(E_i\right)=\frac{1}{13}(i=0,1,2,3,...,12)$
$P\left(\frac{A}{E_i}\right)=0\text{for}i=0,1,2,3$
$P\left(\frac{A}{E_i}\right)=\frac{{}^i{C}_4}{{}^{12}{C}_4}\text{for}i\ge 4$

$a.P\left(A\right)=\sum _{i=0}^{12}P\left(E_i\right)P\left(\frac{A}{E_i}\right)$
$=\frac{1}{13}\times \frac{1}{{}^{12}{C}_4}[{}^4{C}_4+{}^5{C}_4+...+{}^{12}{C}_4]$
$=\frac{{}^{13}{C}_5}{13\times {}^{12}{C}_4}$
$=\frac{1}{5}$

$b.P\left(\frac{A}{E_{10}}\right)=\frac{{}^{10}{C}_4}{{}^{12}{C}_4}=\frac{14}{33}$

$c.P\left(\frac{E_{10}}{A}\right)=\frac{P\left(E_{10}\right)P\left(\frac{A}{E_{10}}\right)}{P\left(A\right)}$
$=\frac{\frac{1}{13}\times \frac{14}{33}}{\frac{1}{5}}$
$=\frac{70}{429}$