SOLUTION: Color blindness appears in 2% of the people in a certain population. How large must a random sample be in order to be 99% certain that a color-blind person is included in the sampl

Question 1178065: Color blindness appears in 2% of the people in a certain population. How large must a random sample be in order to be 99% certain that a color-blind person is included in the sample?
Thank you :)
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**Understanding the Problem**
* **Probability of Color Blindness:** p = 0.02
* **Probability of Not Color Blindness:** q = 1 - p = 0.98
* **Confidence Level:** 99%
* **Goal:** Find the sample size (n) such that we are 99% certain that at least one person in the sample is color blind.
**Approach**
It's easier to calculate the probability that *no one* in the sample is color blind and then find the sample size that makes this probability very small (1 - 0.99 = 0.01).
**Calculations**
1. **Probability of No Color Blindness in a Sample of Size n:**
* The probability that the first person is not color blind is 0.98.
* The probability that the first two people are not color blind is (0.98)².
* The probability that no one in a sample of size n is color blind is (0.98)^n.
2. **Setting Up the Inequality:**
* We want the probability of no color blindness to be less than 1 - 0.99 = 0.01.
* (0.98)^n ≤ 0.01
3. **Solving for n:**
* Take the natural logarithm (ln) of both sides:
* ln((0.98)^n) ≤ ln(0.01)
* n * ln(0.98) ≤ ln(0.01)
* Divide by ln(0.98). Note that ln(0.98) is negative, so we need to reverse the inequality sign:
* n ≥ ln(0.01) / ln(0.98)
* Calculate the logarithms:
* ln(0.01) ≈ -4.60517
* ln(0.98) ≈ -0.0202027
* Calculate n:
* n ≥ -4.60517 / -0.0202027 ≈ 227.94
4. **Round Up:**
* Since n must be an integer, we round up to the nearest whole number.
* n ≥ 228
**Conclusion**
A random sample must be at least 228 people in order to be 99% certain that a color-blind person is included in the sample.

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