Question

A box contains 10 red balls and 15 blue balls. One ball from the box is lost. Two balls are then drawn from the box and are found to be red ones. Find the chance that the missing ball is a blue ball.

• A. $\frac{15}{23}$
• B. 0.347
• C. 0.6
• D. None of these

Answer 1

The correct option is A

$\frac{15}{23}$

Using Baye’s theorem

P(Lost ball is blue/ 2 balls drawn are red)=$P\left(\frac{B}{A_1}\right)\times \frac{P\left(A\right)}{\left[P\right(\frac{B}{A_1})\times P(A_1)+P(\frac{B}{A_2})\times P(A_2\left)\right]}$

$P\left(X\right)=\left(\frac{{}^{10}{C}_2}{{}^{24}{C}_2}\right)\times \frac{\frac{15}{25}}{\left[\right(\frac{{}^{10}{C}_2}{{}^{24}{C}_2})\times \frac{15}{25}+(\frac{{}^9{C}_2}{{}^{24}{C}_2})\times \frac{10}{25}]}$

$P\left(X\right)=\frac{15}{23}$