KCET · Maths · Matrices
If \(A, B, C\) are three mutually exclusive and exhaustive events of an experiment such that \(P(A)=2 P(B)=3 P(C)\), then \(P(B)\) is equal to
- A \(\frac{1}{11}\)
- B \(\frac{2}{11}\)
- C \(\frac{3}{11}\)
- D \(\frac{4}{11}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{11}\)
Step-by-step Solution
Detailed explanation
\(A, B, C\) are mutually exclusive and exhaustive events.
\(\begin{aligned}
\therefore \quad P(A \cap B)=P(B \cap C) \\
=P(A \cap C)=0 &=P(A \cap B \cap C) \\
\text { and } P(A)+P(B)+P(C) &=1 \\
2 P(B)+P(B)+\frac{2}{3} P(B) &=1 \\
{[} &\because P(A)=2 P(B), 2 P(B)=3 P(C)] \\
&=\frac{11 P(B)}{3}=1 \\
P(B) &=\frac{3}{11}
\end{aligned}\)
\(\begin{aligned}
\therefore \quad P(A \cap B)=P(B \cap C) \\
=P(A \cap C)=0 &=P(A \cap B \cap C) \\
\text { and } P(A)+P(B)+P(C) &=1 \\
2 P(B)+P(B)+\frac{2}{3} P(B) &=1 \\
{[} &\because P(A)=2 P(B), 2 P(B)=3 P(C)] \\
&=\frac{11 P(B)}{3}=1 \\
P(B) &=\frac{3}{11}
\end{aligned}\)
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