Respuesta :
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.
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios 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 [tex]\pi[/tex] is the probability of X happening.
In this problem, we have that:
It is reported in about one child in every 500, so [tex]\pi = \frac{1}{500} = 0.002[/tex].
A young obstetrician celebrates her first 100 deliveries, so [tex]n = 100[/tex]
What is the probability that the obstetrician has delivered no child with polydactyly?
That is P(X = 0)
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.002)^{0}.(0.998)^{100} = 0.8186[/tex]
There is a 81.86% probability that the obstetrician has delivered no child with polydactyly.
Using the principle of binomial probability, the probability that the obstetrician delivers no child with polydactyl is : 0.8186
Recall :
[tex]P(X = x) = nCx \times p^{x} \times q^{n-x} [/tex]
- p = probability of success = 1 / 500 = 0.002
- q = 1 - p = 1 - 0.002 = 0.998
- n = number of deliveries = 100
Probability of delivering no child with polydactyl ; x = 0
Substituting the values into the equation
[tex]P(X = 0) = 100C0 \times 0.002^{0} \times 0.998^{100} [/tex]
[tex]P(X = 0) = 100 \times 1 \times 0.8185668 [/tex]
[tex]P(X = x) = 81.85668 [/tex]
Therefore, the probability that the doctor delivers no child with polydactyl is 0.8186
Learn more : https://brainly.com/question/12474772
