Answer:
(A) Probability that all three have type A positive blood is 0.0298.
(B) Probability that none of the three have type A positive blood is 0.3285.
(C) Probability that at least one of the three have A positive blood is 0.6715.
(D) The events which can be considered unusual is of the first part (A).
Explanation:
We are given that the probability that a person in the United States has type A positive blood is 31%.
Three unrelated people in the United states are selected at random.
The above situation can be represented through binomial distribution;
where, n = number of trials (samples) taken = 3 people
r = number of success
p = probability of success which in our question is probability that
a person in the United States has type A positive blood, i.e; 31%
Let X = Number of people in the United States has type A positive blood
So, X ~ Binom(n = 3, p = 0.31)
(A) Probability that all three have type A positive blood is given by = P(X = 3)
P(X = 3) =
=
= 0.0298
(B) Probability that none of the three have type A positive blood is given by = P(X = 0)
P(X = 0) =
=
= 0.3285
(C) Probability that at least one of the three have A positive blood is given by = P(X
1)
P(X
1) = 1 - P(X = 0)
= 1 - 0.3285
= 0.6715
(D) The events above which can be considered unusual is of the first part (A) whose probability is less than 5% because an event is considered unusual whose probability is less than 5%.
Since, the probability of all three people having type A positive blood is 2.98%, so this event is considered to be unusual.