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The probability of getting a success in a trail is five times that of a failure. The probability of getting at most one success in 5 trails, is
Options:
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1989 Upvotes
Verified Answer
The correct answer is:
$\frac{26}{6^5}$
Let the probability of success and failure are $\mathrm{P}$ and $\mathrm{q}$ respectively then, $p=5 q$
$$
\frac{\mathrm{P}}{\mathrm{q}}=\frac{5}{1}
$$
So, $\mathrm{p}=\frac{5}{6}, \mathrm{q}=\frac{1}{6}$
$$
\begin{aligned}
& \mathrm{p}(\text { atmost } \mathrm{t} \text { one success) } \\
& =5_{\angle 0}\left(\frac{5}{6}\right)^0\left(\frac{1}{6}\right)^5+5_{L_1}\left(\frac{5}{6}\right)^1\left(\frac{1}{6}\right)^4 \\
& =\left(\frac{1}{6}\right)^4\left[\frac{1}{6}+3 \times \frac{5}{6}\right]=\left(\frac{1}{6}\right)^4\left[\frac{26}{6}\right]=\frac{26}{6^5}
\end{aligned}
$$
So, option (b) is correct.
$$
\frac{\mathrm{P}}{\mathrm{q}}=\frac{5}{1}
$$
So, $\mathrm{p}=\frac{5}{6}, \mathrm{q}=\frac{1}{6}$
$$
\begin{aligned}
& \mathrm{p}(\text { atmost } \mathrm{t} \text { one success) } \\
& =5_{\angle 0}\left(\frac{5}{6}\right)^0\left(\frac{1}{6}\right)^5+5_{L_1}\left(\frac{5}{6}\right)^1\left(\frac{1}{6}\right)^4 \\
& =\left(\frac{1}{6}\right)^4\left[\frac{1}{6}+3 \times \frac{5}{6}\right]=\left(\frac{1}{6}\right)^4\left[\frac{26}{6}\right]=\frac{26}{6^5}
\end{aligned}
$$
So, option (b) is correct.
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