MHT CET · Maths · Probability
If \(A\) and \(B\) are independent events and \(P(A)=\frac{2}{3}, P(B)=\frac{3}{5}\), then \(P\left(A^{\prime} \cap B\right)=\)
- A \(\frac{4}{15}\)
- B \(\frac{3}{5}\)
- C \(\frac{2}{5}\)
- D \(\frac{1}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{5}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}(\mathrm{A})=\frac{3}{5}, \mathrm{P}(\mathrm{B})=\frac{2}{3}\) and \(\mathrm{A}, \mathrm{B}\) are independent events.
\(
\begin{aligned}
\therefore P(A \cap B) &=P(A) \times P(B)=\frac{3}{5} \times \frac{2}{3}=\frac{2}{5} \\
P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\
&=\frac{3}{5}+\frac{2}{3}-\frac{2}{5}=\frac{13}{15} \\
\therefore P(A \cup B)^{\prime} &=1-\frac{13}{15}=\frac{2}{15} \\
P\left(A^{\prime} \cap B^{\prime}\right) &=P(A \cup B)^{\prime}=\frac{2}{15}
\end{aligned}
\)
\(
\begin{aligned}
\therefore P(A \cap B) &=P(A) \times P(B)=\frac{3}{5} \times \frac{2}{3}=\frac{2}{5} \\
P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\
&=\frac{3}{5}+\frac{2}{3}-\frac{2}{5}=\frac{13}{15} \\
\therefore P(A \cup B)^{\prime} &=1-\frac{13}{15}=\frac{2}{15} \\
P\left(A^{\prime} \cap B^{\prime}\right) &=P(A \cup B)^{\prime}=\frac{2}{15}
\end{aligned}
\)
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