Question

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

Box	Marble colour
	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, then match the terms of column I with their respective values in column II and choose the correct option from the codes given below.

Column I		Column II
A	Probability that it was drawn from box $A$	1	$\frac{8}{15}$
B	Probability that it was drawn	2	$\frac{1}{15}$
C	Probability that it was drawn from box $C$	3	$\frac{2}{5}$

• A. A= 2 ,B= 3 ,C= 1
• B. A= 1 ,B= 2,C=3
• C. A= 3 ,B= 1 ,C=2
• D. A=2 ,B= 3 ,C=1

Answer 1

Answer: (A)

Let $E_1$ : the event that box $A$ is selected, $E_2$ : the event that box $B$ is selected,
$E_3$ : the event that box $C$ is selected and $E_4$ : the event that box $D$ is selected,
$\therefore E_1, E_2, E_3, E_4$ are mutually exclusive and exhaustive. Moreover,
$P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=P\left(E_4\right)=\frac{1}{4}$
Let $E$ : ball drawn is red
$ \therefore P\left(\frac{E}{E_1}\right) =\frac{1}{1+6+3}=\frac{1}{10}$
$\Rightarrow P\left(\frac{E}{E_2}\right)=\frac{6}{6+2+2}=\frac{6}{10} $
$\text { and } P\left(\frac{E}{E_3}\right) =\frac{8}{8+1+1}=\frac{8}{10} $
$P\left(\frac{E}{E_4}\right) =\frac{0}{0+6+4}=0$
By using Bayes' theorem
A. $P($ ball is drawn from box $A)=P\left(\frac{E_1}{E}\right)$
$=\frac{P\left(\frac{E}{E_1}\right) P\left(E_1\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) P\left(E_2\right)+P\left(\frac{E}{E_3}\right) P\left(E_3\right)+P\left(\frac{E}{E_4}\right) P\left(E_4\right)}$
$=\frac{\frac{1}{10} \times \frac{1}{4}}{\frac{1}{10} \times \frac{1}{4}+\frac{6}{10} \times \frac{1}{4}+\frac{8}{10} \times \frac{1}{4}+0 \times \frac{1}{4}}$
$=\frac{\frac{1}{40}}{\frac{1}{4}\left(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0\right)}=\frac{\frac{1}{40}}{\frac{1}{4} \times \frac{15}{10}}=\frac{\frac{1}{40}}{\frac{15}{40}}=\frac{1}{15}$
B. $P$ (ball is drawn from box $B$ )
$=P\left(\frac{E_2}{E}\right)=\frac{P\left(\frac{E}{E_2}\right) P\left(E_2\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) P\left(E_2\right)+P\left(\frac{E}{E_3}\right) P\left(E_3\right)+P\left(\frac{E}{E_4}\right) P\left(E_4\right)}$
$=\frac{\frac{6}{10} \times \frac{1}{4}}{\frac{1}{10} \times \frac{1}{4}+\frac{6}{10} \times \frac{1}{4}+\frac{8}{10} \times \frac{1}{4}+0 \times \frac{1}{4}}$
$=\frac{\frac{6}{40}}{\frac{1}{4}\left(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0\right)}$
$=\frac{\frac{6}{40}}{\frac{1}{4} \times \frac{15}{10}}=\frac{\frac{6}{40}}{\frac{15}{40}}=\frac{6}{40} \times \frac{40}{15}=\frac{2}{5}$
C. $P$ (ball is drawn from box $C$ )
$=P\left(\frac{E_3}{E}\right)=\frac{P\left(\frac{E}{E_3}\right) P\left(E_3\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) P\left(E_2\right)+P\left(\frac{E}{E_3}\right) P\left(E_3\right) +P\left(\frac{E}{E_4}\right) P\left(E_4\right)}$
$=\frac{\frac{8}{10} \times \frac{1}{4}}{\frac{1}{10} \times \frac{1}{4}+\frac{6}{10} \times \frac{1}{4}+\frac{8}{10} \times \frac{1}{4}+0 \times \frac{1}{4}}$
$=\frac{\frac{8}{40}}{\frac{1}{4}\left(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0\right)}$
$=\frac{\frac{8}{40}}{\frac{1}{4} \times \frac{15}{10}}=\frac{8}{\frac{40}{15}}$
$=\frac{8}{40} \times \frac{40}{15}=\frac{8}{15}$