Question

Polydactyly is a fairly common congenital abnormality in which a baby is born with one or more extra fingers or toes. It is reported in about one child in every 500 . A young obstetrician celebrates her first 100 deliveries. What is the probability that the obstetrician has delivered no child with polydactyly? (Enter your answer rounded to four decimal places.)

Answer 1

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.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

It is reported in about one child in every 500, so $\pi = \frac{1}{500} = 0.002$.

A young obstetrician celebrates her first 100 deliveries, so $n = 100$

What is the probability that the obstetrician has delivered no child with polydactyly?

That is P(X = 0)

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{100,0}.(0.002)^{0}.(0.998)^{100} = 0.8186$

There is a 81.86% probability that the obstetrician has delivered no child with polydactyly.