A bag $X$ contains 2 white and 3 black balls and another bag $Y$ contains 4 white and 2 black balls. One bag…
A bag $X$ contains 2 white and 3 black balls and another bag $Y$ contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, the probability for the ball chosen be white, is :
$\frac{2}{15}$
$\frac{7}{15}$
$\frac{8}{15}$
$\frac{14}{15}$
Solution
Probability of selecting a white ball from $X$ bag
$=\frac{2}{5}$
Probability of selecting a white ball from $Y$ bag
$=\frac{4}{6}=\frac{2}{3}$
Probability of selecting a white ball from $X$ or $Y$ bags
$=\frac{2}{5}+\frac{2}{3}=\frac{6+10}{15}=\frac{16}{15}$
Probability of selecting the white ball from one of the bags
$=\frac{1}{2} \cdot \frac{16}{15}=\frac{8}{15}$