It is observed that there will be 25 blood sp | vedclass

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Question

(A) The probability of selecting a normal specimen from a pool of $100$ blood samples is $p = \frac{25}{100} = \frac{1}{4}$.Thus,$q = 1 - p = \frac{3}

Vedclass · Accepted Answer

$1-\frac{13}{4}\left(\frac{3}{4}\right)^9$

Vedclass · Answer

(A) The probability of selecting a normal specimen from a pool of $100$ blood samples is $p = \frac{25}{100} = \frac{1}{4}$.Thus,$q = 1 - p = \frac{3}{4}$.Number of trials $n = 10$.We need to find the probability of having at least two normal specimens,$P(X \geq 2)$.$P(X \geq 2) = 1 - P(X Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^k q^{n-k}$:$P(X = 0) = {}^{10}C_{0} (\frac{1}{4})^0 (\frac{3}{4})^{10} = (\frac{3}{4})^{10}$.$P(X = 1) = {}^{10}C_{1} (\frac{1}{4})^1 (\frac{3}{4})^9 = 10 	imes \frac{1}{4} 	imes (\frac{3}{4})^9 = \frac{10}{4} (\frac{3}{4})^9$.$P(X Therefore,$P(X \geq 2) = 1 - \frac{13}{4} (\frac{3}{4})^9$.

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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