Question

The odds against a certain event is $5:2$ and the odds in favor of another event is $6:5$. If both the events are independent, then find the probability that at least one of the events will happen.

Answer 1

Let probability of the first event taking place be A and probability of the second event taking place be B.

Then,

$\begin{array}{cc}P\left(A\right)& =\frac{2}{5+2}\\ & =2/7\\ P\left(B\right)& =\frac{6}{6+5}\\ & =6/11\end{array}$

The required event can be defined as that $A$ takes place and $B$ does not take place ($A$ or $B$ takes place and $A$ does not take place or $A$ takes place and $B$ takes place.)

$=\left[P\right(A\left)\right\{1-P\left(B\right)\left\}\right]+\left[P\right(B\left)\right\{1-P\left(A\right)\left\}\right]+\left[P\right(A\left)P\right(B\left)\right]$

$=[\frac{2}{7}\times \frac{5}{11}]+[\frac{6}{11}\times \frac{3}{7}]+[\frac{2}{7}\times \frac{6}{11}]$

$=\frac{10+18+12}{77}$

$=\frac{\text{40}}{\text{77}}$