If $A, B, C$ are mutually exclusive and exhaustive events such that $P (A) = 2 P (B) = 3 P (C)$ , then $P (B \cup C) =$

6 / 11

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5 / 11

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4 / 11

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7 / 11

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Solution

The correct option is B $5 / 11$
$P (A) + P (B) + P (C) = 1$

$P (A) + \frac{P (A)}{2} + \frac{P (A)}{3} = 1 \Rightarrow P (A) = \frac{6}{11}$

$∴ P (B) = \frac{3}{11}, P (C) = \frac{2}{11}$

$P (B \cup C) = P (B) + P (C) = \frac{5}{11}$

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Similar questions

A, B

C

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

P (B)

\frac{3}{2}

\frac{1}{2} P (B)

If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 3P(A) = 2P(B) = P(C), then P(A) is equal to

3

A, B

C

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

P (A) =

P (A) = \frac{1}{4}

P (B) = \frac{1}{6}

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

P (A \cup B \cup C) = P (A) + P (B) + P (C)

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