Question

A dice is weighted so that the probability of different faces to turn up is as given

Number	1	2	3	4	5	6
Probability	0.2	0.1	0.1	0.3	0.1	0.2

If P(A/B) = $p_1$, and P(B/C) $=p_2$ and P(C/A) = $p_3$, then the values of $p_1,p_2,p_3$ respectively are -
Take the events A, B and C as $A=\{1,2,3\},B=\{2,3,5\}\text{and}C=\{2,4,6\}$

• A. $\frac{2}{3},\frac{1}{3},\frac{1}{4}$
• B. $\frac{1}{3},\frac{1}{3},\frac{1}{6}$
• C. $\frac{1}{4},\frac{1}{3},\frac{1}{6}$
• D. $\frac{2}{3},\frac{1}{6},\frac{1}{4}$

Answer 1

The correct option is D $\frac{2}{3},\frac{1}{6},\frac{1}{4}$

No.	1	2	3	4	5	6
Prob.	0.2	0.1	0.1	0.3	0.1	0.2

$A=\{1,2,3\},B=\{2,3,5\},C=\{2,4,6\}$

$P_1=P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$

$=\frac{P\left(\right\{2,3\left\}\right)}{P\left(\right\{2,3,5\left\}\right)}$

$=\frac{0.1+0.1}{0.1+0.1+0.1}$

$=\frac{0.2}{0.3}$

$=\frac{2}{3}$

$=P_1$

$P_2=P\left(B/C\right)=\frac{P(B\cap C)}{P\left(C\right)}$

$=\frac{P\left(\right\{2\left\}\right)}{P\left(\right\{2,4,6\left\}\right)}$

$=\frac{0.1}{0.1+0.3+0.2}$

$=\frac{0.1}{0.6}$

$=\frac{1}{6}$

$=P_2$

$P_3=P\left(C/A\right)=\frac{P(C\cap A)}{P\left(A\right)}$

$=\frac{P\left(\right\{2\left\}\right)}{P\left(\right\{1,2,3\left\}\right)}$

$=\frac{0.1}{0.2+0.1+0.1}$

$=\frac{1}{4}$

$P_3=\frac{1}{4}$.