An experiment yields 3 mutually exclusive and exhaustive events A,B and C . If P(A)=2P(.B.)=3P(C), then P(A) is equal to

Question

An experiment yields 3 mutually exclusive and exhaustive events A,B and C . If P(A)=2P(.B.)=3P(C), then P(A) is equal to (A) (1/11) (B) (2/11) (C) (3/11) (D) (6/11)

Tardigrade Admin · Accepted Answer

Given, events A, B and C are mutually exclusive and exhaustive, hence P(A ∪ B ∪ C)=1 ⇒ P(A)+P(B)+P(C)=1 ⇒ P(A)+(1/2)P(A)+(1/3)P(A)=1 ⇒ (11/6)P(A)=1 ⇒ P(A)=(6/11)

Q. An experiment yields $3$ mutually exclusive and exhaustive events $A, B$ and $C$ $.$ If $P (A) = 2 P (B) = 3 P (C),$ then $P (A)$ is equal to

$\frac{1}{11}$

$\frac{2}{11}$

$\frac{3}{11}$

$\frac{6}{11}$

Given, events $A, B$ and $C$ are mutually exclusive and exhaustive, hence $P (A \cup B \cup C) = 1$
$\Rightarrow P (A) + P (B) + P (C) = 1$
$\Rightarrow P (A) + \frac{1}{2} P (A) + \frac{1}{3} P (A) = 1$
$\Rightarrow \frac{11}{6} P (A) = 1$
$\Rightarrow P (A) = \frac{6}{11}$

Q. An experiment yields 3 mutually exclusive and exhaustive events A,B and C . If P(A)=2P(B)=3P(C), then P(A) is equal to

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