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14.Probability
medium
The odds against a certain event is $5 : 2$ and the odds in favour of another event is $6 : 5$. If both the events are independent, then the probability that at least one of the events will happen is
A
$\frac{{50}}{{77}}$
B
$\frac{{52}}{{77}}$
C
$\frac{{25}}{{88}}$
D
$\frac{{63}}{{88}}$
Solution
(b) Let $A$ and $B$ be two given events.
The odds against $A$ are $5:2$, therefore $P(A) = \frac{2}{7}$.
The odds in favour of $B$ are $6:5$, therefore $P(B) = \frac{6}{{11}}.$
The required probability $ = 1 – P(\bar A)\,P(\bar B)$
$ = 1 – \left( {1 – \frac{2}{7}} \right)\,\left( {1 – \frac{6}{{11}}} \right) = \frac{{52}}{{77}}.$
Standard 11
Mathematics
