Question

The chance of an event happening is the square of the chance of a second event but the odds against the first are the cubes of the odds against the second. The chance of happening of each event are

• A. $\frac{1}{9},\frac{1}{3}$
• B. $\frac{1}{6},\frac{1}{9}$
• C. $\frac{1}{3},\frac{1}{6}$
• D. none of these

Answer 1

The correct option is A $\frac{1}{9},\frac{1}{3}$

Let the chance of the second event be $p.$ Then the chance of the first event is $p^2.$
$\therefore$ Odds against the first event are as $1-p^2:p^2.$
and odds against the second event are $1-p:p$
Hence according to the condition given in the question, we have $\frac{1-p^2}{p^2}={\left(\frac{1-p}{p}\right)}^3$
or $\frac{(1-p)(1+p)}{p^2}=\frac{{(1-p)}^3}{p^3}$
or $P(p+1)={(1-p)}^2$
or $p^2+p=p^2-2p+1$ or $3p=1$
$\therefore p=\frac{1}{3}$ and $p^2=\frac{1}{9}$
Hence the probability of the first event $=\frac{1}{9}$
and the probability of the second event $\frac{1}{3}.$