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It is observed that there will be 25 blood specimens of normal persons, if 100 blood samples are tested. If 10 specimens are sent to a laboratory for testing, then the probability of having at least two specimens of normal persons is
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Verified Answer
The correct answer is:
$1-\frac{13}{4}\left(\frac{3}{4}\right)^9$
The probability of selecting a normal specimen from a pool of 100 blood samples is $\frac{25}{100}=\frac{1}{4}$
$$
\therefore \mathrm{P}=\frac{1}{4}
$$
$$
q=1-\frac{1}{4}=\frac{3}{4}
$$
No. of trials $=\mathrm{n}=10$
$$
\begin{aligned}
& P(x \geq 2)=1-P(x < 2) \\
& =1-P(x=1)-P(x=0) \\
& =1-{ }^{10} C_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^9-{ }^{10} C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{10} \\
& =1-\frac{13}{4}\left(\frac{3}{9}\right)^9 \\
& \therefore P(x \geq 2)=1-\frac{13}{4}\left(\frac{3}{9}\right)^9
\end{aligned}
$$
$$
\therefore \mathrm{P}=\frac{1}{4}
$$
$$
q=1-\frac{1}{4}=\frac{3}{4}
$$
No. of trials $=\mathrm{n}=10$
$$
\begin{aligned}
& P(x \geq 2)=1-P(x < 2) \\
& =1-P(x=1)-P(x=0) \\
& =1-{ }^{10} C_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^9-{ }^{10} C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{10} \\
& =1-\frac{13}{4}\left(\frac{3}{9}\right)^9 \\
& \therefore P(x \geq 2)=1-\frac{13}{4}\left(\frac{3}{9}\right)^9
\end{aligned}
$$
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