Question

In a competition A, B and C are participating.The probability that A wins is twice that of B, the probability that B wins is twice that of C. Then the probability that A loses is

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{4}{7}$
• D. $\frac{3}{7}$

Answer 1

The correct option is D $\frac{3}{7}$

$P\left(A\right)=2P\left(B\right)$
$P\left(B\right)=2P\left(C\right)$
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$
A, B and C are mutually exclusive and exhaustive events.
$1=P\left(A\right)+\frac{P\left(A\right)}{2}+\frac{P\left(A\right)}{4}$
$P\left(A\right)=\frac{4}{7}$
Then the probability that A loses is: $1-P\left(A\right)=1-\frac{4}{7}=\frac{3}{7}$