Question

$A,B$ and $C$ are three mutually exclusive and exhaustive events such that $P\left(A\right)=2P\left(B\right)=3P\left(C\right)$.
What is $P\left(B\right)$?

• A. $6/11$
• B. $6/22$
• C. $1/6$
• D. $1/4$

Answer 1

Answer: (D)

$1/4$

Given: A, B, C are mutually exclusive and exhaustive events, so this means

$P(A\cup B\cup C)=P\left(S\right)=1$

And ,$P(A\cap B\cap C)=P(A\cap B)=P(B\cap C)=P(A\cap C)=0$

We know,

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)⟹1=P\left(A\right)+P\left(B\right)+P\left(C\right)-0-0-0+0⟹P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

Give, $P\left(A\right)=2P\left(B\right)=2P\left(C\right)$

Hence, $P\left(A\right)+P\left(B\right)+P\left(C\right)=1$ becomes

$2P\left(B\right)+P\left(B\right)+P\left(B\right)=1⟹4P\left(B\right)=1⟹P\left(B\right)=\frac{1}{4}$