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Probability

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Question : 125 of 127

Marks: +1, -0

A,

B

C

are three mutually exclusive and exhaustive events such that

P(A)=2 P(B)=3 P(C)

P(B)?

$\frac{6}{11}$
$\frac{3}{11}$
$\frac{1}{6}$
$\frac{1}{3}$

Solution:

Concept:

Use the condition of mutual exclusivity and exhaustiveness to sum probabilities to 1, expressing all in terms of

P(B)

.

Explanation:

Let

P(B) = x

.

Given

P(A) = 2 P(B) = 2x

and

P(A) = 3 P(C)

, so

2x = 3 P(C) \Rightarrow P(C) = \frac{2}{3}x

.

Since events are mutually exclusive and exhaustive,

P(A) + P(B) + P(C) = 1

.

Thus

2x + x + \frac{2}{3}x = 1

.

Combine:

\left(3 + \frac{2}{3}\right)x = \frac{11}{3}x = 1

.

Solve:

x = \frac{3}{11}

.

Therefore

P(B) = \frac{3}{11}

.

Answer:

\frac{3}{11}

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