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  • If P(A) = 2|5, P(B) = 3|10 and P(A ∩ B) = 1|5, the P(A′|B′).P(B′|A′) is equal to A. 5|6 B. 5|7 C. 25|42 D. 1

If P(A)=\frac{2}{5}, P(B)=\frac{3}{10} \text { and } P(A \cap B)=\frac{1}{5}, the P(A′|B′).P(B′|A′) is equal to
A. 5|6
B. 5|7
C. 25|42
D. 1

 

Answers (1)

Given- 

\begin{aligned} &P(A)=\frac{2}{5}, P(B)=\frac{3}{10} \text { and } P(A \cap B)=\frac{1}{5}\\ &\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)\\ &\Rightarrow P\left(A^{\prime} \mid B^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}\\ &\mathrm{P}\left(\mathrm{B}^{\prime} \mid \mathrm{A}^{\prime}\right) \times \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\mathrm{P}\left(\mathrm{B}^{\prime} \cap \mathrm{A}^{\prime}\right)\\ &\Rightarrow P\left(B^{\prime} \mid A^{\prime}\right)=\frac{P\left(B^{\prime} \cap A^{\prime}\right)}{P\left(A^{\prime}\right)}\\ &\text { Multiplying eq. (i) and (ii), we get }\\ \end{aligned}

\begin{aligned} &P\left(A^{\prime} \mid B^{\prime}\right) \times P\left(B^{\prime} \mid A^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)} \times \frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(A^{\prime}\right)}\\ &=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)} \times \frac{1-P(A \cup B)}{P\left(A^{\prime}\right)}\\ &\left[\because P\left(A^{\prime} \cap B^{\prime}\right)=P\left[(A \cup B)^{\prime}\right]=1-P(A \cup B)\right]\\ &=\frac{(1-P(A \cup B))^{2}}{(1-P(B)) \times(1-P(A))}\end{aligned}

 

\\ =\frac{\left(1-(P(A)+P(B)-P(A \cap B))^{2}\right.}{\left(1-\frac{3}{10}\right)\left(1-\frac{2}{5}\right)} \\ =\frac{\left[1-\left(\frac{2}{5}+\frac{3}{10}-\frac{1}{5}\right)\right]^{2}}{\left(\frac{10-3}{10}\right)\left(\frac{5-2}{5}\right)} \\ =\frac{\left[1-\left(\frac{4+3-2}{10}\right)\right]^{2}}{\frac{7}{10} \times \frac{3}{5}}

\\ =\frac{\left[1-\left(\frac{5}{10}\right)\right]^{2}}{\frac{7}{10} \times \frac{3}{5}} \\ =\frac{\left[1-\frac{1}{2}\right]^{2}}{\frac{7}{10} \times \frac{3}{5}} \\ =\frac{\left[\frac{1}{2}\right]^{2}}{\frac{7}{10} \times \frac{3}{5}} \\ =\frac{1}{4} \times \frac{50}{21} \\ =\frac{25}{42}

Hence, the correct option is C

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