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Q.
A and B agree to play 6 games of pool. Assume that the probability of A wins a game is p where $p$ is fixed and $0
Probability - Part 2
Solution:
${ }^6 C _4 p ^4(1- p )^2={ }^6 C _5 p ^5(1- p )$
$\Rightarrow 15(1- p )=6 p$
$\therefore \frac{1-p}{p}=\frac{2}{5}$
Now, $\frac{{ }^6 C _2 p ^2(1- p )^4}{{ }^6 C _3 p ^3(1- p )^3}=\frac{15}{20} \times \frac{(1- p )}{ p }=\frac{3}{4} \times \frac{2}{5}=\frac{3}{10}$