Question

A person with type AB blood can donate red blood cells to people with the same blood type. About 12% of the US population has type AB blood. University High held a blood drive where 50 students donated blood. Part A: What is the probability that exactly 8 of the students had type AB blood? Part B: What is the probability that at least 8 of the students had type AB blood?

Answer 1

Using the binomial distribution, it is found that there is a:

A. 0.1075 = 10.75% probability that exactly 8 of the students had type AB blood.

B. 0.2467 = 24.67% probability that at least 8 of the students had type AB blood.

For each student, there are only two possible outcomes, either they have type AB blood, or they do not. The probability of a student having type AB blood is independent of any other student, hence the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• About 12% of the US population has type AB blood, hence $p = 0.12$.

• A sample of 50 students is taken, hence $n = 50$.

Item a:

The probability is P(X = 8), hence:

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

$P(X = 8) = C_{50,8}.(0.12)^{8}.(0.88)^{42} = 0.1075$

0.1075 = 10.75% probability that exactly 8 of the students had type AB blood.

Item b:

The probability is:

$P(X \geq 8) = 1 - P(X < 8)$

In which:

$P(X < 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

Using a calculator to find each probability, we have that:

$P(X < 8) = 0.7533$

Then:

$P(X \geq 8) = 1 - P(X < 8) = 1 - 0.7533 = 0.2467$

0.2467 = 24.67% probability that at least 8 of the students had type AB blood.

You can learn more about the binomial distribution at https://brainly.com/question/24863377