Question

Children inherit their blood type from their parents, with probabilities that reflect the parents' genetic makeup. Children of Juan and Maria each have probability 1/4 of having blood type A and inherit independently of each other. Juan and Maria plan to have 4 children; let X be the number who have blood type A. (a) What are n and p in the binomial distribution of X

Answer 1

`n = 4, p = 0.25`

Explanation:

We are given the following in the question:

We treat adult having blood type A as a success.

P(Blood type A) =

`p=(1)/(4) = 0.25`

Then the number of Children of Juan and Maria who have blood type A follows a binomial distribution as there are

• n independent trial.
• Each trial have to outcomes(Blood type A or not)
• Probability of success is same for each success

Here,

`P(X=x) = \binom{n}{x}.p^x.(1-p)^(n-x)`

where n is the total number of observations, x is the number of success, p is the probability of success.

Thus, for the given binomial distribution,

`n = 4, p = 0.25`