Question

The odds against the player A to win are $5:2$ and odds in favor of another player to win are $6:5$. If the two events are independent, Find the probability that at least one player will win.

• A. $\frac{52}{77}$
• B. $\frac{25}{77}$
• C. $\frac{7}{11}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

$\frac{52}{77}$

Given odds against A to win are $5:2$
So, odds in favour of A are $2:5$
$P\left(A\right)=\frac{2}{5+2}=\frac{2}{7}$
$\Rightarrow P\left(\overline{A}\right)=\frac{5}{7}$
Odds in favor of B to win are $6:5$
$P\left(B\right)=\frac{6}{11}$
$\Rightarrow P\left(\overline{B}\right)=\frac{5}{11}$
Probability of not winning of A and B both is $P\left(\overline{A}\right).P\left(\overline{B}\right)$
So, the probability of winning of at least one is
$=1-P\left(\overline{A}\right).P\left(\overline{B}\right)$
$=1-\frac{5}{7}\frac{5}{11}$
$=\frac{52}{77}$