Answer:
We know that:
1/5 = 0.2 of all adults in US have type O blood.
Then: 4/5 = 0.8 of all adults in US do not have type O blood.
a) All four are type O.
Let's think this, as four events with two possible outcomes, having type O blood, with a probability of 0.2, and not having type O blood with a probability of 0.8
Now if we want the four people to have type O blood, then we have the outcome with probability 0.2 four times, and as we know, the joint probability is equal to the product of the individual probabilities:
P =0.2^4 = 0.0016
b) None of them is type O.
Similar to the above case, but this time we work with four times the probability 0.8
P = 0.8^4 = 04096
c) Two of the four are type O.
Here we will have two times the probability 0.2, and two times the probability 0.8.
p = 0.2^2*0.8^2
But, if the people is ordered, the two persons with type O blood can be:
first and second
first and third
first and fourth.
second and third
second and fourth
third and fourth.
6 different combinations, then the actual probability for this case will be:
P = 6*p = 6*(0.2^2*0.8^2) = 0.1536
