Question

Box I contains $2$ white and $3$ red balls and box II contains $4$ white and $5$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?

• A. $\frac{54}{44}$
• B. $\frac{54}{14}$
• C. $\frac{54}{104}$
• D. None of these

Answer 1

Answer: (C)

$\frac{54}{104}$

Probability that the ball drawn is red and from ! =$P\left(R/A\right)$

$P\left(R/A\right)=\frac{P\left(A/R\right)\times P\left(A\right)}{P\left(B/R\right)\times P\left(B\right)+P\left(A/R\right)\times P\left(A\right)}$

$P\left(R/A\right)=\frac{3}{5},P\left(R/B\right)=\frac{5}{9}$

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{2}$

$P\left(R/A\right)=\frac{\frac{3}{5}\times \frac{1}{2}}{\frac{5}{9}\times \frac{1}{2}+\frac{3}{5}\times \frac{1}{2}}=\frac{54}{104}$