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 ABY: the number of Australians with blood type ABWhat is the probability that exactly 2 of the U.S. residents have blood type AB? (Note: Some answers are rounded)Group of answer choices0.13340.09880.26460.040.2762
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 ABY: the number of Australians with blood type ABWhat is the probability that exactly 2 of the U.S. residents have blood type AB? (Note: Some answers are rounded)Group of answer choices0.13340.09880.26460.040.2762
Solution
The probability of an event can be calculated using the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability we are trying to find,
- C(n, k) is the combination of n items taken k at a time,
- p is the probability of success (in this case, the probability of a person having blood type AB),
- n is the number of trials (in this case, the number of people sampled), and
- k is the number of successes we want (in this case, the number of people with blood type AB).
For this problem, we want to find the probability that exactly 2 of the 50 U.S. residents have blood type AB. So, we have:
- n = 50 (the number of U.S. residents sampled),
- k = 2 (the number of U.S. residents we want to have blood type AB), and
- p = 0.04 (the probability of a U.S. resident having blood type AB, given as 4%).
Substituting these values into the binomial probability formula, we get:
P(X=2) = C(50, 2) * (0.04^2) * ((1-0.04)^(50-2))
Calculating this gives a probability of approximately 0.2762. So, the probability that exactly 2 of the U.S. residents have blood type AB is approximately 0.2762, or 27.62%.
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