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

Question:

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

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?, box C?

Solution:

Let R be the event of drawing the red marble.

Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C.

Total number of marbles = 40

Number of red marbles = 15

$\therefore P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing the red marble from box $A$ is given by $P\left(E_{A} \mid R\right)$.

$\therefore P\left(E_{A} \mid R\right)=\frac{P\left(E_{A} \cap R\right)}{P(R)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}$

Probability that the red marble is from box $B$ is $P\left(E_{B} \mid R\right)$.

$\Rightarrow P\left(E_{B} \mid R\right)=\frac{P\left(E_{B} \cap R\right)}{P(R)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}$

Probability that the red marble is from box $C$ is $P\left(E_{C} \mid R\right)$.

$\Rightarrow P\left(E_{C} \mid R\right)=\frac{P\left(E_{C} \cap R\right)}{P(R)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}$