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 1 child in every 500. A young obstetrician celebrates her first 100 deliveries.

What is the probability that she has delivered no child with polydactyly? (Think about the chance that a particular individual child delivered does not have polydactyly, and work from there)

What is the probability that she has delivered at least one child with polydactyly?

81.86% probability that she has delivered no child with polydactyly

18.14% probability that she has delivered at least one child with polydactyly.

Step-by-step explanation:

For each delivery, there are only two possible outcomes. Either the baby has polydactyly, or he does not. The probabilitis of babies having polydactyly are independent from each other. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

It is reported in about 1 child in every 500.

This means that

A young obstetrician celebrates her first 100 deliveries.

This means that

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

This is

81.86% probability that she has delivered no child with polydactyly

What is the probability that she has delivered at least one child with polydactyly?

Either she has delivered no child with polydactyly, or she has delivered at least one. The sum of the probabilities of these events is decimal 1. So

We want

So

18.14% probability that she has delivered at least one child with polydactyly.

there is no grid

step-by-step explanation:

570, the answer is 570, hope it