Probability Question 399
Question: 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
Options:
A) $ \frac{1}{3} $
B) $ \frac{1}{9} $
C) $ \frac{2}{3} $
D) $ \frac{4}{9} $
Show Answer
Answer:
Correct Answer: B
Solution:
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}={{( \frac{1-q}{q} )}^{3}}\Rightarrow \frac{1-q^{2}}{q^{2}}=\frac{{{(1-q)}^{3}}}{q^{3}} $
[From (i)]
$ \Rightarrow ( \frac{1-q}{q^{2}} )[ (1+q)-\frac{{{(1-q)}^{2}}}{q} ]=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} $
BETA
