Question

The probability that a person in the United States has type B​+ blood is 12​%. Three unrelated people in the United States are selected at random. Complete parts​ (a) through​ (d). ​(a) Find the probability that all three have type B​+ blood. The probability that all three have type B​+ blood is nothing. ​(Round to six decimal places as​ needed.)

Answer 1

Answer:

The probability that all three have type B​+ blood is 0.001728

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have type B+ blood, or they do not. The probability of a person having type B+ blood is independent of any other person. So we use the binomial probability distribution to solve this question.

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 combinations 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.

The probability that a person in the United States has type B​+ blood is 12​%.

This means that $p = 0.12$

Three unrelated people in the United States are selected at random.

This means that $n = 3$

Find the probability that all three have type B​+ blood.

This is P(X = 3).

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

$P(X = 3) = C_{3,3}.(0.12)^{3}.(0.88)^{0} = 0.001728$

The probability that all three have type B​+ blood is 0.001728