About 1/5 of all adults in the United States have type O blood. If four randomly selected adults donate blood, find the probability of each of the following events. a. All four are type O.
b. None of them is type O.
c. Two out of the four are type O .

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

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70