Question

A bag contains $15$ red and $20$ black balls. Each ball is numbered $1,2\text{or}3$. $20\%$ of red balls are numbered $1$ and $40\%$ are numbered $3$. Similarly $45\%$ of black balls are numbered $2$ and $30\%$ are numbered $3$. One balls is drawn at random and found to be numbered $2$, then probability that it was red ball is

Answer 1

(1) (2) (3)
Red balls $20\%$ $40\%$ $40\%$
Black balls $25\%$ $45\%$ $30\%$
$p=\frac{\frac{15}{35}\times \frac{40}{100}}{\frac{15}{35}\times \frac{40}{100}+\frac{20}{35}\times \frac{45}{100}}=\frac{40}{40+60}=\frac{2}{5}$