A,B,C are mutually exclusive and exhaustive event associated with random experiment . Find p(A)
P(B)=3/2 p(A) and p(C)=1/2p(B)

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Solution

If A,B,C are mutually exclusive and exhaustive events
P(A)+P(B)+P(C)=1
P(B)=3/2P(A)
P(C)=1/2P(B)=1/2×3/2P(A)
=3/4P(A)
so eqn become
P(A)+3/2P(A)+3/4P(A)=1
=P(A)[1+3/2+3/4]=1
=P(A)[1+9/4]=1
= P(A)[13/4]=1
P(A)=1/13/4=4/13

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Similar questions

A, B, C are three mutually exclusive and exhaustive events associated with a random experiment.

If $P (B) = (\frac{3}{2}) P (A)$ and $P (C) = (\frac{1}{2}) P (B),$ find P(A).

P (A)

P (B) = \frac{3}{2} P (A)

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

\frac{3}{2}

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

A, B

C

P (B) = \frac{3}{2} P (A)

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

P (A \cup C) =

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