Question

A bag contains $3$ red, $2$ white and $2$ black balls. Two balls are drawn at random and none of them is found to be a white ball. The probability that both balls are red is $\frac{a}{b}$ (where $a,b$ are coprime) then $b-a$ is equal to

Answer 1

$E=2$ balls drawn, none of them is white
$E_{1}=$ both balls red
$E_{2}=$ both balls black
$E_{3}=$ one red, one black
$P\left(E_{1} / E\right)=\frac{P \left(E / E_{1}\right) \cdot P \left(E_{1}\right)}{P \left(E / E_{1}\right) \cdot P \left(E_{1}\right) + P \left(E / E_{2}\right) \cdot P \left(E_{2}\right) + P \left(E / E_{3}\right) \cdot P \left(E_{3}\right)}$
$=\frac{\frac{\_{}^{3}C_{2}^{}}{\_{}^{7}C_{2}^{}}}{\frac{\_{}^{3}C_{2}^{}}{\_{}^{7}C_{2}^{}} + \frac{\_{}^{2}C_{2}^{}}{\_{}^{7}C_{2}^{}} + \frac{\_{}^{3}C_{2}^{} \cdot \_{}^{2}C_{1}^{}}{\_{}^{7}C_{2}^{}}}=\frac{\_{}^{3}C_{2}^{}}{\_{}^{3}C_{2}^{} + \_{}^{2}C_{2}^{} + \_{}^{3}C_{1}^{} \cdot \_{}^{2}C_{1}^{}}$
$=\frac{3}{3 + 1 + 3 . 2}=\frac{3}{10}$