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A person fails 4 times in a game when he plays 9 times. If he plays 15 times, the probability of having success at most one is
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Verified Answer
The correct answer is:
$\frac{79}{9}\left(\frac{4}{9}\right)^{14}$
We have, $P=\frac{5}{9}$
$$
\therefore \quad q=1-p=1-\frac{5}{9}=\frac{4}{9}
$$
and
$$
n=15, r \leq 1
$$
$$
\begin{aligned}
\therefore \quad P(x \leq 1) & =P(X=0)+P(X=1) \\
& ={ }^{15} C_0\left(\frac{5}{9}\right)^0\left(\frac{4}{9}\right)^{15}+{ }^{15} C_{\mathrm{r}}\left(\frac{5}{9}\right)^1\left(\frac{4}{9}\right)^{151} \\
& =\left(\frac{4}{9}\right)^{15}+15 \times \frac{5}{9} \times\left(\frac{4}{9}\right)^{14} \\
& =\left(\frac{4}{9}\right)^{14}\left[\frac{4}{9}+\frac{75}{9}\right] \\
& =\frac{79}{9}\left(\frac{4}{9}\right)^{14}
\end{aligned}
$$
$$
\therefore \quad q=1-p=1-\frac{5}{9}=\frac{4}{9}
$$
and
$$
n=15, r \leq 1
$$
$$
\begin{aligned}
\therefore \quad P(x \leq 1) & =P(X=0)+P(X=1) \\
& ={ }^{15} C_0\left(\frac{5}{9}\right)^0\left(\frac{4}{9}\right)^{15}+{ }^{15} C_{\mathrm{r}}\left(\frac{5}{9}\right)^1\left(\frac{4}{9}\right)^{151} \\
& =\left(\frac{4}{9}\right)^{15}+15 \times \frac{5}{9} \times\left(\frac{4}{9}\right)^{14} \\
& =\left(\frac{4}{9}\right)^{14}\left[\frac{4}{9}+\frac{75}{9}\right] \\
& =\frac{79}{9}\left(\frac{4}{9}\right)^{14}
\end{aligned}
$$
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