KCET · Maths · Trigonometric Ratios & Identities
The probability of solving a problem by three persons \(A, B\) and \(C\) independently is \(\frac{1}{2}, \frac{1}{4}\) and \(\frac{1}{3}\) respectively. Then the probability of the problem is solved by any two of them is
- A \(\frac{1}{12}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{24}\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Given, \(P(A)=\frac{1}{2}, P(B)=\frac{1}{4}, P(C)=\frac{1}{3}\)
\(P(\bar{A})=\frac{1}{2}, P(\bar{B})=\frac{3}{4}, P(\bar{C})=\frac{2}{3}\)
Required probability \(=P(A B \bar{C})+P(A \bar{B} C)+P(\bar{A} B C)\)
\(\begin{aligned}
&=\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{3}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3} \\
&=\frac{2+3+1}{24}=\frac{6}{24}=\frac{1}{4}
\end{aligned}\)
\(P(\bar{A})=\frac{1}{2}, P(\bar{B})=\frac{3}{4}, P(\bar{C})=\frac{2}{3}\)
Required probability \(=P(A B \bar{C})+P(A \bar{B} C)+P(\bar{A} B C)\)
\(\begin{aligned}
&=\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{3}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3} \\
&=\frac{2+3+1}{24}=\frac{6}{24}=\frac{1}{4}
\end{aligned}\)
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