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In an experiment a person gets success $\alpha$ times out of $\beta$ trails. If the experiment consists of $n$ trials, then the probability that he fails at least $(n-1)$ times is
Options:
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Verified Answer
The correct answer is:
$\frac{(\beta-\alpha)^{n-1}}{\beta^n}(n \alpha+\beta-\alpha)$
Probability for success $p=\frac{\alpha}{\beta}$
Probability for failure $q=\frac{\beta-\alpha}{\beta}$
Apply binomial theorem,
Probability of atleast fails $(n-1)$ times means pass the exam at most one time.
$\mathrm{P}$ (fails atleast $(n-1)$ times)
$$
={ }^n C_0\left(\frac{\alpha}{\beta}\right)\left(\frac{\beta-\alpha}{\beta}\right)^n+{ }^n C_1\left(\frac{\alpha}{\beta}\right)^1\left(\frac{\beta-\alpha}{\beta}\right)^{n-1}
$$
$$
\begin{aligned}
& =\frac{(\beta-\alpha)^{n-1}}{\beta^n}\left[\frac{\underline{n}}{\underline{n}}(\beta-\alpha)+\frac{n \underline{n-1}}{\lfloor n-1} \alpha\right] \\
& =\frac{(\beta-\alpha)^{n-1}}{(\beta)^n}[\beta-\alpha+n \alpha]
\end{aligned}
$$
So, option (b) is correct.
Probability for failure $q=\frac{\beta-\alpha}{\beta}$
Apply binomial theorem,
Probability of atleast fails $(n-1)$ times means pass the exam at most one time.
$\mathrm{P}$ (fails atleast $(n-1)$ times)
$$
={ }^n C_0\left(\frac{\alpha}{\beta}\right)\left(\frac{\beta-\alpha}{\beta}\right)^n+{ }^n C_1\left(\frac{\alpha}{\beta}\right)^1\left(\frac{\beta-\alpha}{\beta}\right)^{n-1}
$$
$$
\begin{aligned}
& =\frac{(\beta-\alpha)^{n-1}}{\beta^n}\left[\frac{\underline{n}}{\underline{n}}(\beta-\alpha)+\frac{n \underline{n-1}}{\lfloor n-1} \alpha\right] \\
& =\frac{(\beta-\alpha)^{n-1}}{(\beta)^n}[\beta-\alpha+n \alpha]
\end{aligned}
$$
So, option (b) is correct.
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