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The odds against a certain event are $5: 2$ and the odds in favour of another independent event are $6: 5 .$ The probability that at least one of the events will happen, is
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Verified Answer
The correct answer is:
$\frac{52}{77}$
Let $A$ and $B$ are two given events.
$$
\begin{array}{l}
\therefore \quad P(A)=\frac{2}{7}, P(B)=\frac{6}{11} \\
\therefore \text { Required probability }=1-P(\bar{A}) P(\bar{B}) \\
\quad=1-\left(1-\frac{2}{7}\right)\left(1-\frac{6}{11}\right) \\
=\frac{52}{77}
\end{array}
$$
$$
\begin{array}{l}
\therefore \quad P(A)=\frac{2}{7}, P(B)=\frac{6}{11} \\
\therefore \text { Required probability }=1-P(\bar{A}) P(\bar{B}) \\
\quad=1-\left(1-\frac{2}{7}\right)\left(1-\frac{6}{11}\right) \\
=\frac{52}{77}
\end{array}
$$
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