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:a. X is a binomial random variable with n = 50 and p = 0.04b. Y is a binomial random variable with n = 40 and p = 0.015Step-by-step explanation:Binomial probability distributionThe binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]And p is the probability of X happening.X: the number of US residents (out of 50) with blood type AB.Blood type AB is the rarest blood type, occurring in only 4% of the population in the United StatesThis means that [tex]p = 0.04, n = 50[/tex] Y: the number of Australians (out of 40) with blood type AB.In Australia, only 1.5% of the population has blood type AB. This means that [tex]p = 0.015, n = 40[/tex] Z: the total number of individuals (out of 90) with blood type AB.Here[tex]n = 90, p = 0.04*\frac{50}{90} + 0.015*\frac{40}{90} = 0.0289[/tex]Which of the following is true about the random variables X, Y, and Z?Options a and b are true, while c is false.