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Mathematics Question on Probability

Let A and B be two events such that

P(B/A)25, P(A/B)=17 and P(A∩B)=19.Consider(S1)P(A’∪B)=56,\begin{array}{l} P\left(B/A\right)\frac{2}{5},\ P\left(A/B\right)=\frac{1}{7}\ \text{and}\ P\left(A\cap B\right)=\frac{1}{9}.\end{array} Consider \begin{array}{l} \left(S1\right)P\left(A’\cup B\right)=\frac{5}{6},\end{array}

(S2)P(A’∩B’)=118.\begin{array}{l} \left(S2\right)P\left(A’\cap B’\right)=\frac{1}{18}.\end{array}Then

A

Both (S1) and (S2) are true

B

Both (S1) and (S2) are false

C

Only (S1) is true

D

Only (S2) is true

Answer

Both (S1) and (S2) are true

Explanation

Solution

P(A/B)=17⇒P(A∩B)P(B)=17\begin{array}{l} P\left(A/B\right)=\frac{1}{7}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{7}\end{array}

⇒P(B)=79\begin{array}{l} \Rightarrow P\left(B\right)=\frac{7}{9}\end{array}

P(B/A)=25⇒P(A∩B)P(A)=25\begin{array}{l} P\left(B/A\right)=\frac{2}{5}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{2}{5} \end{array}

P(A)=52⋅19=518\begin{array}{l} P\left(A\right)=\frac{5}{2}\cdot \frac{1}{9}=\frac{5}{18}\end{array}

S2:P(A’∩B’)=118\begin{array}{l} S2:P\left(A’\cap B’\right)=\frac{1}{18}\end{array}

S1: and P(A’∪B)=19+69+118=56.\begin{array}{l} S1:\text{ and }P\left(A’\cup B\right)=\frac{1}{9}+\frac{6}{9}+\frac{1}{18}=\frac{5}{6}.\end{array}