Question

Blood type AB is the rarest blood type, occurring in only 4% of the population in the United States. In Australia, only 1.5% of the population has blood type AB. Suppose a random sample of 50 U.S. residents and 40 Australians is obtained. Consider the random variables described below: X: the number of U.S. residents with blood type AB Y: the number of Australians with blood type AB What is the probability that exactly 2 of the U.S. residents have blood type AB? (Note: Some answers are rounded) 0.2762 0.04 0.1334 0.0988 0.2646

Answer 1

Answer:

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

Step-by-step explanation:

For each U.S. resident, there are only two outcomes possible. Either they have blood type AB, or they do not. This means that we can solve this problem using binomial probability distribution concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

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 $\pi$ is the probability of X happening.

In this problem, we have that:

50 U.S residents are sampled, so $n = 50$

4% of the U.S population has blood type AB, so $p = 0.04$.

What is the probability that exactly 2 of the U.S. residents have blood type AB?

This is P(X = 2). So:

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

$P(X = 2) = C_{50,2}.(0.04)^{2}.(0.96)^{48} = 0.2762$

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.