The united states, 44% of the population has type a blood. consider taking a sample of size 4. let y denote the number of persons in the sample with type a blood. find pr[y = 0}. pr{ y=1}. pr{y = 2}.

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%

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logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.