Respuesta :
Answer:
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
It is reported in about 1 child in every 500.
This means that [tex]p = \frac{1}{500} = 0.002[/tex]
A young obstetrician celebrates her first 100 deliveries.
This means that [tex]n = 100[/tex]
What is the probability that she has delivered no child with polydactyly?
This is [tex]P(X = 0)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.002)^{0}.(0.998)^{100} = 0.8186[/tex]
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
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want [tex]P(X \geq 1)[/tex]
So
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8186 = 0.1814[/tex]
18.14% probability that she has delivered at least one child with polydactyly.
