In a large population of 10,000 lab mice, each mouse has an independent 0.0003 probability of carrying a rare genetic mutation.
(B) Use the Poisson distribution to estimate the probability that 2 mice carry the mutation.

• A. $0.78$
• B. $0.22$
• C. $0.18$
• D. $0.82$

Answer: (B)

Verified step by step guidance

1. Step 1: Recognize that the problem involves a rare event (probability of 0.0003 per mouse) in a large population (10,000 mice). This is a classic scenario where the Poisson distribution is a good approximation for the binomial distribution.
2. Step 2: Calculate the expected number of mice carrying the mutation (λ), which is the mean of the Poisson distribution. Use the formula λ = n × p, where n is the population size (10,000) and p is the probability of carrying the mutation (0.0003).
3. Step 3: Write the Poisson probability formula: P(X = k) = (λ^k * e^(-λ)) / k!, where k is the number of successes (in this case, 2 mice carrying the mutation), λ is the mean, and e is the base of the natural logarithm (approximately 2.718).
4. Step 4: Substitute the values into the formula. Use λ calculated in Step 2 and k = 2. Compute the numerator (λ^k * e^(-λ)) and the denominator (k!).
5. Step 5: Simplify the expression to find the probability P(X = 2). This will give the estimated probability that exactly 2 mice carry the mutation using the Poisson distribution.