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Probability

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Question : 54 of 127

Marks: +1, -0

A, Band

C

are three mutually exclusive and exhaustive events such that

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

P(B) ?

$\frac{6}{11}$
$\frac{6}{22}$
$\frac{1}{6}$
$\frac{1}{3}$

Solution:

Concept:

Use the property that for mutually exclusive and exhaustive events, the sum of probabilities is

1

.

Explanation:

Let

P(B) = x

.

Given

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

and

2P(B) = 3P(C)

, so

2x = 3P(C)

\Rightarrow

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

.

Since

A

,

B

,

C

are mutually exclusive and exhaustive:

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

.

Thus

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

.

Combine:

\frac{11x}{3} = 1

\Rightarrow

x = \frac{3}{11}

.

So

P(B) = \frac{3}{11} = \frac{6}{22}

.

Answer:

\frac{6}{22}

(Option B).

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