Question

According to the Greater New York Blood Program, 45% of donors have a group O blood type. Construct a probability distribution to show the probabilities that, in a group of 5 donors, 0, 1, 2, 3, 4, or 5 of them have a group O blood type. (Round your probabilities to three decimal places.)

a. Probability of 0

b. Probability of 1

c. Probability of 2

d. Probability of 3

e. Probability of 4

Answer 1

To create a probability distribution for the number of donors with type O blood in a group of five, the binomial probability formula is used, where the probability of type O blood is 45%.

Step-by-step explanation:

Constructing a probability distribution for the number of donors in a group of 5 that have a group O blood type, given that 45% of donors have this blood type, involves using the binomial probability formula:

P(X = k) = C(n, k) × (p)^k × (1-p)^(n-k)

Where:

• C(n, k) is the combination of n items taken k at a time
• p is the probability of success (having type O blood)
• n is the total number of trials (donors)
• k is the number of successes among those trials.

Here, p = 0.45, n = 5.

• Probability of 0: P(0) = C(5, 0) × (0.45)^0 × (0.55)^5 = 1 × 1 × 0.55^5 ≈ 0.050
• Probability of 1: P(1) = C(5, 1) × (0.45)^1 × (0.55)^4 = 5 × 0.45 × 0.55^4 ≈ 0.250
• Probability of 2: P(2) = C(5, 2) × (0.45)^2 × (0.55)^3 ≈ 0.365
• Probability of 3: P(3) = C(5, 3) × (0.45)^3 × (0.55)^2 ≈ 0.234
• Probability of 4: P(4) = C(5, 4) × (0.45)^4 × (0.55)^1 ≈ 0.086
• Probability of 5: P(5) = C(5, 5) × (0.45)^5 × (0.55)^0 ≈ 0.015

Each probability is rounded to three decimal places, as requested.