A person with type A blood can donate red blood cells to people with type A or type AB blood. About 31% of the US population has type A blood. University High held a blood drive where 50 students donated blood.

Part A: What is the probability that exactly 17 of the students had type A blood? (5 points)

Part B: What is the probability that at least 17 of the students had type A blood? (5 points)

A) The probability that exactly 17 of the students had type A blood is; P(X = 17) = 0.1068

B) The probability that at least 17 of the students had type A blood is; P(X ≤ 17) = 0.7334

This is a binomial probability question that can be solved with the formula;

P(X = x) = ⁿCₓ × pˣ × (1 - p)ⁿ ⁻ ˣ

We are given;

p = 31% = 0.31

n = 50

A) Probability that exactly 17 of the students had type A blood is;

P(X = 17) = ⁵⁰C₁₇ 0.31¹⁷ (1 - 0.31)⁽⁵⁰ ⁻ ¹⁷)

P(X = 17) = 0.1068

B) The probability that at least 17 of the students had type A blood is;

P(X ≤ 17) = P(X = 17) + P(X = 16) + P(X = 15) + P(X = 14) + P(X = 0)

Using binomial probability calculator, we have;

P(X ≤ 17) = 0.7334

Read more about binomial probability distribution at; link

0.107

0.733

Step-by-step explanation:

Part A) Use binomial probability.

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

n = 50, r = 17, p = 0.31, q = 0.69.

P = ₅₀C₁₇ 0.31¹⁷ 0.69⁵⁰⁻¹⁷

P ≈ 0.107

Or, using binompdf function in a calculator:

binompdf(n, p, r)

= binompdf(50, 0.31, 17)

≈ 0.107

Part B) This time, you'll need to use the binomcdf function in a calculator.

binomcdf(n, p, r)

= binomcdf(50, 0.31, 17)

≈ 0.733

sites:

$6.42

step-by-step explanation: