The proportions of blood phenotypes, A, B, AB, and O, in the population of all Caucasians in the United States are approximately .41, .10, .04, and .45, respectively. A single Caucasian is chosen at random from the population. a List the sample space for this experiment. b Make use of the information given above to assign probabilities to each of the simple events. c What is the probability that the person chosen at random has either type A or type AB blood

a. S = {A,B,AB,O}

b. P(A) = 0.41, P(B) = 0.1, P(AB) = 0.04, P(O) = 0.45

c. 0.45 = 45% probability that the person chosen at random has either type A or type AB blood

Step-by-step explanation:

a List the sample space for this experiment.

All possible blood types(A, B, AB or O). So

S = {A,B,AB,O}

b Make use of the information given above to assign probabilities to each of the simple events.

The proportions of blood phenotypes, A, B, AB, and O, in the population of all Caucasians in the United States are approximately .41, .10, .04, and .45, respectively.

So

P(A) = 0.41, P(B) = 0.1, P(AB) = 0.04, P(O) = 0.45

c What is the probability that the person chosen at random has either type A or type AB blood

P(A) + P(AB) = 0.41 + 0.04 = 0.45

0.45 = 45% probability that the person chosen at random has either type A or type AB blood

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.

the probability that a bird is green (.30) is true

step-by-step explanation: