Answer:
There is a 81.86% probability that the obstetrician has delivered no child with polydactyly.
Step-by-step explanation:
There are only two possible outcomes: Either the baby has the anormality, or he hasn't. So we use the binomial probability distribution.
Binomial probability
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.%5Cpi%5E%7Bx%7D.%281-%5Cpi%29%5E%7Bn-x%7D)
In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And
is the probability of X happening.
In this problem, we have that:
It is reported in about one child in every 500, so
.
A young obstetrician celebrates her first 100 deliveries, so ![n = 100](https://tex.z-dn.net/?f=n%20%3D%20100)
What is the probability that the obstetrician has delivered no child with polydactyly?
That is P(X = 0)
![P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.%5Cpi%5E%7Bx%7D.%281-%5Cpi%29%5E%7Bn-x%7D)
![P(X = 0) = C_{100,0}.(0.002)^{0}.(0.998)^{100} = 0.8186](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B100%2C0%7D.%280.002%29%5E%7B0%7D.%280.998%29%5E%7B100%7D%20%3D%200.8186)
There is a 81.86% probability that the obstetrician has delivered no child with polydactyly.