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Mathematics Question on Conditional Probability

Let AA and BB be two event such that P(A)=38,P(B)=58P\left(A\right) = \frac{3}{8}, P\left(B\right) = \frac{5}{8} and P(A∪B)=34P\left(A\cup B\right) =\frac{3}{4}. then P(A∣B)⋅P(A′∣B)P \left(A|B\right)\cdot P\left(A'|B\right) is equal to

A

25\frac{2}{5}

B

38\frac{3}{8}

C

320\frac{3}{20}

D

625\frac{6}{25}

Answer

625\frac{6}{25}

Explanation

Solution

P(A)=38,P(B)=58,P(A∪B)=34P\left(A\right) = \frac{3}{8}, P\left(B\right) =\frac{5}{8}, P \left(A\cup B\right)= \frac{3}{4} P(A∩B)=P(A)+P(B)−P(A∪B)P \left(A\cap B\right) = P \left(A\right) +P\left(B\right) -P\left(A\cup B\right) =38+58−34= \frac{3}{8} +\frac{5}{8} - \frac{3}{4} =3+5−68=14= \frac{3+5-6}{8} = \frac{1}{4} We know that P(A′∩B)+P(A∩B)=P(B)P\left(A' \cap B\right) + P\left(A \cap B\right) = P\left(B\right) [As A′∩BA '\cap B and A∩BA \cap B are mutually exclusive events] P(A′∩B)=P(B)−P(A∩B)P\left(A'\cap B\right)= P\left(B\right) - P\left(A\cap B\right) =58−14=38= \frac{5}{8} -\frac{1}{4} = \frac{3}{8} Now, P(A∣B)⋅P(A′∣B)=P(A∩B)p(B)⋅P(A′∩B)P(B)P\left(A|B\right)\cdot P\left(A'|B\right) =\frac{ P\left(A\cap B\right)}{p\left(B\right)} \cdot\frac{ P\left(A'\cap B\right)}{P\left(B\right)} =1/45/8⋅3/85/8= \frac{1/4}{5/8}\cdot\frac{{3}/{8}}{{5}/{8} } =332×6425=625 = \frac{3}{32} \times\frac{64}{25} = \frac{6}{25}