Question

In an essay competition, tho odds is favour of competitors $P,Q,R,S$ are $1:2,1:3,1:4$, and $1:5$ respectively. Find the probability that one of them wins the competition.

Answer 1

Probability that P wins $=\frac{1}{2}$

Probability that Q wins $=\frac{1}{3}$

Probability that R wins $=\frac{1}{4}$

Probability that S wins $=\frac{1}{5}$

Thus, Probability that P loss $=1-\frac{1}{2}=\frac{1}{2}$

Probability that Q loss $=1-\frac{1}{3}=\frac{2}{3}$

Probability that R loss $=1-\frac{1}{4}=\frac{3}{4}$

Probability that S loss $=1-\frac{1}{5}=\frac{4}{5}$

Probability that one of them wins the competitions

$=(P\times Q^{\prime }\times R^{\prime }\times S^{\prime })+(P^{\prime }\times Q\times R^{\prime }\times S^{\prime })+(P^{\prime }\times Q^{\prime }\times R\times S^{\prime })+(P^{\prime }\times Q^{\prime }\times R^{\prime }\times S)$

$=(\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5})+(\frac{1}{3}\times \frac{1}{2}\times \frac{3}{4}\times \frac{4}{5})+(\frac{1}{4}\times \frac{1}{2}\times \frac{2}{3}\times \frac{4}{5})+(\frac{1}{5}\times \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4})$

$=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}+\frac{1}{20}$

$=\frac{(12+6+4+3)}{60}$

$=\frac{25}{60}$

$=\frac{5}{12}$