Answer:
pr{Y=0} = 9,8%
pr{Y=1} = 30.9%
pr{Y=2} = 36%
Step-by-step explanation:
The probability of each person in the sample having type A blood in independent from each other. So, the first person will have a 44% chance of having type A blood, and so will the second, the third and the fourth.
So, the first question.
a) pr{Y = 0}
The probability of each of the four people having non-type A blood is 56%.
So, pr{Y=0} = 0.56*0.56*0.56*0.56 = 0.098 = 9,8%.
b) pr{Y=1}
All of the following order of samples satisfy the condition
A - NA - NA - NA
NA - A - NA - NA
NA - NA - A - NA
NA - NA - NA - A
Since the probabilties are independent, each order has the following probability P.
P = 0.44*(0.56)^{3} = 0.0773
Since there are four possible orders, pr{Y=1} = 4*P = 0.309 = 30.9%
c) pr{Y=2}
All of the following order of samples satisfy the condition:
A - A - NA - NA
A - NA - A - NA
A - NA - NA - A
NA - A - A - NA
NA - A - NA - A
NA - NA - A - A
Each order has the following probability P.
P = (0.44)²*(0.56)² = 0.06
Since there are six possible orders, pr{Y = 2} = 6*P = 0.36 = 36%