Question

the probabilities of four mutually exclusive and exhaustive events a,b,c and d, satisfy the relation : 2p(a) = 3p(b) = 4p(c) = 6p(d). the probability of event c is :​

Answer 1

Answer:

$P(C)=\frac{3}{14}$

Step-by-step explanation:

Concept used:

Two events A and B are said to be mutually exclusive events if

$A\cap\:B=\phi$

Two events A and b are said to be mutually exhaustive events if

$A\cup\:B=S$

Given:

$2\:P(A)=3\:P(B)=4\:P(C)=6\:P(D)=k(say)$

Then,

$P(A)=\frac{k}{2}$

$P(B)=\frac{k}{3}$

$P(C)=\frac{k}{4}$

$P(D)=\frac{k}{6}$

since A,B,C and D are mutually exclusive and exhaustive events,

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

$\frac{k}{2}+\frac{k}{3}+\frac{k}{4}+\frac{k}{6}=1$

$\frac{6k+4k+3k+k}{12}=1$

$\frac{14k}{12}=1$

$\frac{7k}{6}=1$

$k=\frac{6}{7}$

Now,

$P(C)$

$=\frac{k}{4}$

$=\frac{\frac{6}{7}}{4}$

$=\frac{6}{28}$

$=\frac{3}{14}$

$P(C)=\frac{3}{14}$