Answer:
the probability that the obstetrician has delivered no child with polydactyly in her first 100 deliveries = 0.819
Step-by-step explanation:
This is a binomial distribution problem
Probability of a polydactyl case, p = 1/500 = 0.002
Probability of no polydactyl case, q = 1 - 0.002 = 0.998
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
For this case,
x = 0
n = 100
P(X = 0) = ¹⁰⁰C₀ (0.02)⁰ (0.998)¹⁰⁰⁻⁰
P(X = 0) = ¹⁰⁰C₀ (0.02)⁰ (0.998)¹⁰⁰
P(X = 0) = 1 × 1 × 0.819 = 0.819