Respuesta :
Answer:
3% conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy
Step-by-step explanation:
For each children, there are only two possible outcomes. Either they are allergic to peanuts, or they are not. The probability of a children being allergic to peanuts is independent of other children. So we use the binomial probability distribution to solve this question.
For the final question, the conditional probability formula is used.
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.
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this problem:
Event A: At least one person allergic to peanuts.
Event B: Exactly two children are allergic to peanuts.
P(A)
3% of children in the United States are allergic to peanuts.
This means that [tex]p = 0.03[/tex]
3 children, so [tex]n = 3[/tex]
Either no children is allergic, or at least one is. The sum of these probabilities is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We have that [tex]P(A) = P(X \geq 1)[/tex]
So
[tex]P(A) = P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.03)^{0}.(0.97)^{3} = 0.9127[/tex]
[tex]P(A) = P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9127 = 0.0873[/tex]
Intersection:
Intersection of at least one children and exactly two, so exactly two.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(A \cap B) = P(X = 2) = C_{3,2}.(0.03)^{2}.(0.97)^{1} = 0.002619[/tex]
Probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.002619}{0.0873} = 0.03[/tex]
3% conditional probability that exactly two of the children will be allergic to peanuts, given that at least one of the three children suffers from this allergy
