Question

An experiment resulting in sample space as S={a,b,c,d,e,f} with $P\left(a\right)=\frac{1}{16},$ $P\left(b\right)=\frac{1}{16},$ $P\left(c\right)=\frac{2}{16},$ $P\left(d\right)=\frac{3}{16},$ $P\left(e\right)=\frac{4}{16},$ and $P\left(f\right)=\frac{5}{16},$
Let three events A,B and C are defined as A={a,c,e}, B={c,d,e,f} and C={b,c,f}.
If P(A/B)=$p_1$, P(B/C)=$p_2,$ $P\left(C/A^c\right)=p_3$ and $P\left(A^c/C\right)=p_4,$ then the correct order sequance is

• A. $p_1<p_3<p_2<p_4$
• B. $p_1<p_4<p_3<p_2$
• C. $p_1<p_3<p_4<p_2$
• D. $p_3<p_1<p_4<p_2$

Answer 1

The correct option is C $p_1<p_3<p_4<p_2$

$\begin{array}{c}p\left(a\right)=\frac{1}{16},p\left(b\right)=\frac{1}{16},p\left(c\right)=\frac{2}{16},p\left(d\right)=\frac{3}{16},p\left(e\right)=\frac{4}{16},andp\left(f\right)=\frac{5}{16}\\ A=\left\{a,c,e\right\}\\ p\left(A\right)=p\left(a\right)+p\left(c\right)+p\left(e\right)\\ =\frac{7}{16}\\ B=\left\{c,d,e,f\right\}\\ p\left(B\right)=p\left(c\right)+p\left(d\right)+p\left(e\right)+p\left(f\right)\\ =\frac{14}{16}\\ C=\left\{b,c,f\right\}\\ p\left(c\right)=p\left(b\right)+p\left(c\right)+p\left(f\right)\\ =\frac{8}{16}\\ p_1=p\left(\frac{A}{B}\right)=\frac{p\left(A\cap B\right)}{p\left(B\right)}\\ =\frac{p\left(A\right).p\left(B\right)}{p\left(B\right)}\\ and,\left(A\cap B\right)=\left\{c,e\right\}\\ p\left(A\cap B\right)=\frac{\frac{6}{16}}{\frac{14}{16}}=\frac{3}{7}=p_1\\ p_2=p\left(\frac{B}{C}\right)=\frac{p\left(B\cap C\right)}{p\left(C\right)}\\ \\ and,\left(B\cap C\right)=\left\{c,f\right\}\\ p\left(B\cap C\right)=\frac{7}{16}\\ P_2=\frac{7}{8}\\ P_3=P\left(\frac{C}{A}\right)=\frac{p\left(C\cap A\right)}{p\left(A\right)}\\ \left(C\cap A\right)=\left\{b,f\right\}\\ p\left(C\cap A\right)=\frac{6}{16}\\ p\left(A\right)=\frac{9}{16}\\ p\left(\frac{A}{C}\right)=\frac{6}{9}=p_3\\ p_4=p\left(\frac{A}{C}\right)\\ similary\\ p_4=\frac{6}{16}\\ Hence.\\ p_2>p_3>p_3>p_1\\ Hence,theoptionCisthecorrectanswer.\end{array}$