Question

The blood type of successive children born to the same parents are independent and have fixedprobabilities that depend on the genetic makeup of the parents. Each child born to a particularset of parents has probability 0.15 of having blood type O. If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?

Answer 1

Answer:

There is a 9.75% probability that exactly 2 of them have type O blood.

Step-by-step explanation:

For each children, there is only two possible outcomes. Either they have blood type O, or they do not. This means that 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

4 children, so $n = 4$

Probability 0.15 of having blood type O, so $p = 0.15$

If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.15)^{2}.(0.85)^{2} = 0.0975$

There is a 9.75% probability that exactly 2 of them have type O blood.