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

Question

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<p<1. If the probability of A wins 4 games out of 6 games is equal to the probability that he wins 5 games out of 6 games, then the ratio of the probability that he wins 2 games out of 6 games to the probability that he wins 3 out of 6 games is, (A) (1/5) (B) (3/10) (C) (9/20) (D) (1/2)

Tardigrade Admin · Accepted Answer

6 C 4 p 4(1- p )2= 6 C 5 p 5(1- p ) ⇒ 15(1- p )=6 p ∴ (1-p/p)=(2/5) Now, ( 6 C 2 p 2(1- p )4/ 6 C 3 p 3(1- p )3)=(15/20) × ((1- p )/ p )=(3/4) × (2/5)=(3/10)

$\frac{1}{5}$

$\frac{3}{10}$

$\frac{9}{20}$

$\frac{1}{2}$

$^{6} C_{4} p^{4} (1 - p)^{2} =^{6} C_{5} p^{5} (1 - p)$
$\Rightarrow 15 (1 - p) = 6 p$
$∴ \frac{1 - p}{p} = \frac{2}{5}$
Now, $\frac{^{6} C _{2} p ^{2} ( 1 - p ) ^{4}}{^{6} C _{3} p ^{3} ( 1 - p ) ^{3}} = \frac{15}{20} \times \frac{( 1 - p )}{p} = \frac{3}{4} \times \frac{2}{5} = \frac{3}{10}$

51​

103​

209​

21​

6C4​p4(1−p)2=6C5​p5(1−p) ⇒15(1−p)=6p ∴p1−p​=52​ Now, 6C3​p3(1−p)36C2​p2(1−p)4​=2015​×p(1−p)​=43​×52​=103​

$\frac{1}{5}$

$\frac{3}{10}$

$\frac{9}{20}$

$\frac{1}{2}$

$^{6} C_{4} p^{4} (1 - p)^{2} =^{6} C_{5} p^{5} (1 - p)$
$\Rightarrow 15 (1 - p) = 6 p$
$∴ \frac{1 - p}{p} = \frac{2}{5}$
Now, $\frac{^{6} C _{2} p ^{2} ( 1 - p ) ^{4}}{^{6} C _{3} p ^{3} ( 1 - p ) ^{3}} = \frac{15}{20} \times \frac{( 1 - p )}{p} = \frac{3}{4} \times \frac{2}{5} = \frac{3}{10}$