Suppose we have four boxes $A, B, C$ and $D$ containing coloured marbles as given below:

Box	Red	White	Black
$A$	$1$	$6$	$3$
$B$	$6$	$2$	$2$
$C$	$8$	$1$	$1$
$D$	$0$	$6$	$4$

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red,what is the probability that it was drawn from box $A$,box $B$,or box $C$?

(A) Let $E_A, E_B, E_C,$ and $E_D$ be the events of selecting boxes $A, B, C,$ and $D$ respectively. Since one box is selected at random,$P(E_A) = P(E_B) = P(E_C) = P(E_D) = \frac{1}{4}$.
Let $R$ be the event of drawing a red marble.
The conditional probabilities of drawing a red marble from each box are:
$P(R|E_A) = \frac{1}{1+6+3} = \frac{1}{10}$
$P(R|E_B) = \frac{6}{6+2+2} = \frac{6}{10}$
$P(R|E_C) = \frac{8}{8+1+1} = \frac{8}{10}$
$P(R|E_D) = \frac{0}{0+6+4} = 0$
By the Law of Total Probability,$P(R) = P(E_A)P(R|E_A) + P(E_B)P(R|E_B) + P(E_C)P(R|E_C) + P(E_D)P(R|E_D)$
$P(R) = \frac{1}{4} \times \frac{1}{10} + \frac{1}{4} \times \frac{6}{10} + \frac{1}{4} \times \frac{8}{10} + \frac{1}{4} \times 0 = \frac{1+6+8}{40} = \frac{15}{40} = \frac{3}{8}$.
Using Bayes' Theorem,the probability that the red marble was drawn from box $X$ is $P(E_X|R) = \frac{P(E_X)P(R|E_X)}{P(R)}$.
For box $A$: $P(E_A|R) = \frac{(1/4)(1/10)}{3/8} = \frac{1/40}{3/8} = \frac{1}{40} \times \frac{8}{3} = \frac{1}{15}$.
For box $B$: $P(E_B|R) = \frac{(1/4)(6/10)}{3/8} = \frac{6/40}{3/8} = \frac{6}{40} \times \frac{8}{3} = \frac{2}{5}$.
For box $C$: $P(E_C|R) = \frac{(1/4)(8/10)}{3/8} = \frac{8/40}{3/8} = \frac{8}{40} \times \frac{8}{3} = \frac{8}{15}$.