Question

If A and B are independent events such that
$P\left(A\right)=p,P\left(B\right)=2p$ and
$P\left(\text{Exactly one of A and B}\right)=\frac{5}{9}$
then $p=$

• A. $\frac{1}{3}$
• B. $\frac{5}{6}$
• C. $\frac{5}{12}$
• D. $\frac{9}{11}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{3}$
C $\frac{5}{12}$

$\mathrm{In}\mathrm{set}\mathrm{theory},\mathrm{exactly}\mathrm{one}\mathrm{of}A\mathrm{and}B\mathrm{can}\mathrm{be}\mathrm{denoted}\mathrm{as}\left(A-B\right)\cup \left(B-A\right)\mathrm{or},\left(A\cap B^C\right)\cup \left(B\cap A^C\right)\therefore P\left(\mathrm{Exactly}\mathrm{one}\mathrm{of}A\mathrm{and}B\right)=P\left(\left(A\cap B^C\right)\cup \left(B\cap A^C\right)\right)=P\left(A\cap B^C\right)+P\left(B\cap A^C\right)\left(\because A\cap B^C\mathrm{and}B\cap A^C\mathrm{are}\mathrm{disjoint}.\right]=P\left(A\right).P\left(B^C\right)+P\left(B\right).P\left(A^C\right)\left(\because A\mathrm{and}B\mathrm{are}\mathrm{independent}.\right]\mathrm{Given}P\left(A\right)=p,P\left(B\right)=2p\mathrm{and}P\left(\mathrm{Exactly}\mathrm{one}\mathrm{of}A\mathrm{and}B\right)=\frac{5}{9}\Rightarrow p\left(1-2p\right)+2p\left(1-p\right)=\frac{5}{9}\Rightarrow 3p-4p^2=\frac{5}{9}\Rightarrow p=\frac{1}{3},\frac{5}{12}$