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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When dealing with "at least" probabilities, it's often easier to calculate the probability of the complement event and subtract it from 1. Remember the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k q^{n-k} \).
Step 1: Define the probability of a specimen being from a normal person.
Out of 100 blood samples, 25 are from normal persons. So, the probability that a randomly selected blood specimen is from a normal person is:
\[
p = \frac{25}{100} = \frac{1}{4}
\]
The probability that a randomly selected blood specimen is not from a normal person (i.e., abnormal) is:
\[
q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}
\]
Step 2: Identify the event of interest and its complement.
We are interested in the probability of having at least two specimens of normal persons when 10 specimens are sent for testing. Let \( X \) be the number of specimens from normal persons in the sample of 10. This is a binomial distribution with parameters \( n = 10 \) and \( p = \frac{1}{4} \). We want to find \( P(X \ge 2) \).
It is easier to calculate the probability of the complement event, which is having less than two specimens of normal persons, i.e., \( P(X<2) = P(X = 0) + P(X = 1) \).
Then, \( P(X \ge 2) = 1 - P(X<2) = 1 - [P(X = 0) + P(X = 1)] \).
Step 3: Calculate \( P(X = 0) \) and \( P(X = 1) \) using the binomial probability formula.
The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k q^{n-k} \).
For \( X = 0 \) (no specimens from normal persons):
\[
P(X = 0) = \binom{10}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^{10} = 1 \times 1 \times \left(\frac{3}{4}\right)^{10} = \left(\frac{3}{4}\right)^{10}
\]
For \( X = 1 \) (exactly one specimen from a normal person):
\[
P(X = 1) = \binom{10}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^{10-1} = 10 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^{9} = \frac{10}{4} \left(\frac{3}{4}\right)^{9} = \frac{5}{2} \left(\frac{3}{4}\right)^{9}
\]
Step 4: Calculate \( P(X \ge 2) \).
\[
P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)] = 1 - \left[ \left(\frac{3}{4}\right)^{10} + \frac{5}{2} \left(\frac{3}{4}\right)^{9} \right]
\]
To combine the terms, we can write \( \left(\frac{3}{4}\right)^{10} = \frac{3}{4} \left(\frac{3}{4}\right)^{9} \):
\[
P(X \ge 2) = 1 - \left[ \frac{3}{4} \left(\frac{3}{4}\right)^{9} + \frac{10}{4} \left(\frac{3}{4}\right)^{9} \right] = 1 - \left[ \left( \frac{3}{4} + \frac{10}{4} \right) \left(\frac{3}{4}\right)^{9} \right]
\]
\[
P(X \ge 2) = 1 - \frac{13}{4} \left(\frac{3}{4}\right)^{9}
\]
This matches option (2).
