JEE Main & Advanced Mathematics Probability Question Bank Self Evaluation Test - Probability-I

The chance of one event happening is the square of the chance of a second event, but the odds against the first are the cube of the odds against the second. The chance of the first event is

A) \[\frac{1}{3}\]

B) \[\frac{1}{9}\]

C) \[\frac{2}{3}\]

D) \[\frac{4}{9}\]

Correct Answer: B

Solution :

[b] Let the two events be \[{{E}_{1}}\] and\[{{E}_{2}}\]. Let their chances be p and q respectively.

Then \[p={{q}^{2}}\] ? (i)

The chances of not happening of the events are

Odds against the first event \[=\frac{1-p}{p}\]

Odds against the second event \[=\frac{1-q}{q}\]

Given \[\frac{1-p}{p}={{\left( \frac{1-q}{q} \right)}^{3}}\Rightarrow \frac{1-{{q}^{2}}}{{{q}^{2}}}=\frac{{{(1-q)}^{3}}}{{{q}^{3}}}\]

[From (i)]

\[\Rightarrow \left( \frac{1-q}{{{q}^{2}}} \right)\left[ (1+q)-\frac{{{(1-q)}^{2}}}{q} \right]=0\]

\[\because q\ne 1\] And \[q\ne 0\]

\[\therefore q(1+q)=1-2q+{{q}^{2}}\Rightarrow q=\frac{1}{3}\]

\[\therefore \] from (i) \[\therefore p={{q}^{2}}=\frac{1}{9}\]

\[\therefore p({{E}_{1}})=p=\frac{1}{9}\]