MHT CET · Maths · Probability
The odds against solving a problem by \(A\) and \(B\) are \(3: 2\) and \(2: 1\) respectively, then the probability that the problem will be solved, is
- A \(\frac{3}{5}\)
- B \(\frac{2}{15}\)
- C \(\frac{2}{5}\)
- D \(\frac{11}{15}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{5}\)
Step-by-step Solution
Detailed explanation
Given, \(P(A)=\frac{3}{2+3}=\frac{3}{5}, P(B)=\frac{2}{2+1}=\frac{2}{3}\)
and \(P(\bar{A})=1-\frac{3}{5}=\frac{2}{5}, P(\bar{B})=1-\frac{2}{3}=\frac{1}{3}\)
\(\therefore\) Required probability
\(
\begin{array}{l}
=P(A \cap \bar{B})+P(\bar{A} \cap B)+P(\bar{A} \cap \bar{B}) \\
=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(\bar{A}) \cdot P(\bar{B}) \\
=\frac{3}{5} \cdot \frac{1}{3}+\frac{2}{5} \cdot \frac{2}{3}+\frac{2}{5} \cdot \frac{1}{3} \\
=\frac{1}{5}+\frac{4}{15}+\frac{2}{15} \\
=\frac{3+4+2}{15}=\frac{9}{15}=\frac{3}{5}
\end{array}
\)
and \(P(\bar{A})=1-\frac{3}{5}=\frac{2}{5}, P(\bar{B})=1-\frac{2}{3}=\frac{1}{3}\)
\(\therefore\) Required probability
\(
\begin{array}{l}
=P(A \cap \bar{B})+P(\bar{A} \cap B)+P(\bar{A} \cap \bar{B}) \\
=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(\bar{A}) \cdot P(\bar{B}) \\
=\frac{3}{5} \cdot \frac{1}{3}+\frac{2}{5} \cdot \frac{2}{3}+\frac{2}{5} \cdot \frac{1}{3} \\
=\frac{1}{5}+\frac{4}{15}+\frac{2}{15} \\
=\frac{3+4+2}{15}=\frac{9}{15}=\frac{3}{5}
\end{array}
\)
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