Answer: (a) The probability that all three have type B+ blood is (0.11) * (0.11) * (0.11) = 0.001331 = 0.1331% (rounded to six decimal places).
(b) Find the probability that exactly two of the three have type B+ blood.
The probability that exactly two of the three have type B+ blood can be found using the formula for combinations: C(3,2) * (0.11)^2 * (0.89)^1, where C(3,2) represents the number of ways to choose 2 people out of 3, (0.11)^2 is the probability of both of those people having type B+ blood, and (0.89)^1 is the probability of the third person not having type B+ blood. So, C(3,2) * (0.11)^2 * (0.89)^1 = 3 * 0.11^2 * 0.89 = 0.3027 (rounded to four decimal places).
(c) Find the probability that exactly one of the three has type B+ blood.
The probability that exactly one of the three has type B+ blood can be found using the formula for combinations: C(3,1) * (0.11)^1 * (0.89)^2, where C(3,1) represents the number of ways to choose 1 person out of 3, (0.11)^1 is the probability of that person having type B+ blood, and (0.89)^2 is the probability of the other two people not having type B+ blood. So, C(3,1) * (0.11)^1 * (0.89)^2 = 3 * 0.11 * 0.89^2 = 0.2613 (rounded to four decimal places).
(d) Find the probability that none of the three has type B+ blood.
The probability that none of the three has type B+ blood is (0.89)^3 = 0.696731 = 69.6731% (rounded to four decimal places).
Explanation: