Question 1199567: . In the School of ICT, one person in 80, on average, has blood of type O. If 200 student blood donors are
taken at random, find an approximation to the probability that they include at least five persons having
blood of type O.
How many student donors must be taken at random in order that the probability of including at least one
student donor of type O shall be 0.9 or more
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! To solve this, we approach each part of the problem systematically using probability theory and approximations.
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### **Part 1: Approximation for at least 5 persons with blood type O**
1. **Given data**:
- Probability of having type O blood: \( p = \frac{1}{80} = 0.0125 \)
- Sample size: \( n = 200 \)
- Random variable: Let \( X \) be the number of students with blood type O. \( X \) follows a **binomial distribution**:
\[
X \sim \text{Binomial}(n = 200, p = 0.0125)
\]
2. **Normal approximation**:
Since \( n \) is large and \( p \) is small, we approximate \( X \) using a normal distribution with:
- Mean: \( \mu = n \cdot p = 200 \cdot 0.0125 = 2.5 \)
- Standard deviation: \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{200 \cdot 0.0125 \cdot 0.9875} \approx 1.567 \)
- Approximation: \( X \sim N(\mu = 2.5, \sigma = 1.567) \)
3. **Find \( P(X \geq 5) \):**
Using the normal approximation, apply continuity correction:
\[
P(X \geq 5) \approx P\left(Z \geq \frac{5 - 0.5 - \mu}{\sigma}\right)
\]
Substitute values:
\[
Z = \frac{5 - 0.5 - 2.5}{1.567} = \frac{2}{1.567} \approx 1.276
\]
From standard normal tables or a calculator:
\[
P(Z \geq 1.276) = 1 - P(Z \leq 1.276) \approx 1 - 0.8980 = 0.1020
\]
Thus:
\[
P(X \geq 5) \approx 0.1020
\]
---
### **Part 2: Number of donors for \( P(\text{at least one type O}) \geq 0.9 \)**
1. **Given data**:
- Probability of having type O blood: \( p = 0.0125 \)
- Required: Find \( n \) such that \( P(\text{at least one type O}) \geq 0.9 \).
2. **Complement rule**:
\[
P(\text{at least one type O}) = 1 - P(\text{none with type O})
\]
For \( P(\text{none with type O}) \), the probability is:
\[
P(\text{none with type O}) = (1 - p)^n
\]
Thus:
\[
1 - (1 - p)^n \geq 0.9 \quad \Rightarrow \quad (1 - p)^n \leq 0.1
\]
3. **Solve for \( n \):**
Taking the natural logarithm:
\[
\ln((1 - p)^n) \leq \ln(0.1) \quad \Rightarrow \quad n \cdot \ln(1 - p) \leq \ln(0.1)
\]
Substitute \( p = 0.0125 \):
\[
\ln(1 - 0.0125) \approx \ln(0.9875) \approx -0.0126
\]
\[
n \cdot (-0.0126) \leq \ln(0.1) \approx -2.3026
\]
\[
n \geq \frac{-2.3026}{-0.0126} \approx 183
\]
Thus, at least **183 donors** are required to ensure the probability of including at least one donor with type O blood is \( \geq 0.9 \).
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### **Final Answers**:
1. Probability of at least 5 type O donors: **0.1020**.
2. Number of donors needed for \( P(\text{at least one type O}) \geq 0.9 \): **183**.
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