The American Red Cross says that about 45% of the U. S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. a) Someone volunteers to give blood. What is the probability that this donor 1. has Type AB blood? 2. has Type A or Type B? 3. is not Type O? b) Among four potential donors, what is the probability that 1. all are Type O? 2. no one is Type AB? 3. they are not all Type A? 4. at least one person is Type B?

Question

Mathematics

Reuben

24 June, 06:55

The American Red Cross says that about 45% of the U. S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. a) Someone volunteers to give blood. What is the probability that this donor 1. has Type AB blood? 2. has Type A or Type B? 3. is not Type O? b) Among four potential donors, what is the probability that 1. all are Type O? 2. no one is Type AB? 3. they are not all Type A? 4. at least one person is Type B?

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Terrance Petersen · Answer 1

First part:

1) The probability of this person having type AB blood is 4% because this is the proportion of the U. S. population that has type AB blood. This is calculated by adding the proportion ot the U. S. population that has type O blood with the proportion that has type A blood and the proportion that has type B blood and then finding out the value for us to complete the 100%.

45% + 40% + 11% = 96%

2) In this case we must use the addition rule:

P (A or B) = P (A) + P (B) - P (A and B)

In this case A is the event that the person has type A blood. B is the event that the person has type B blood.

P (A) = 40%

P (B) = 11%

P (A and B) = 0 because you cannot hay type A and type B blood at the same time.

P (A or B) = 51%

3) The probability that the person has not type O blood is going to be the same as the proportion of the U. S. population that doesn't have type O blood.

100% - 45% = 55%

Second part:

In this case we have 4 trials. Each trial is independent so we can use the multiplication rule.

P (A and B and C and D) = P (A) x P (B) * P (c) * P (D)

1) P (O and O and 0 and O) = 0.45 * 0*45*0.45*0.45 = 0.041 = 4.1%

2) 0,96 * 0.96 * 0.96 * 0.96 = 0.85 = 85%

3) In this case we need to add the probability that 1 of them is type B plus the probability that 2 of them are type B plus the probability that 3 of them are type B plus the probability that the tour of them are type B

The probability of being type B is 11% = 0.11

The probability of not being type Bis 89%=0.89

0.11 * 0.89 X 0.89 * 0.89 = 0.078 = 7.8%

0>11 * 0.11 * 0.89*0.89 = 0.0096 = 0.96%

0.11 * 0.11 * 0.11 * 0.89 = 0.0012 = 0.12%

0.11*0.11*0.11*0.11 = 0.00015 = 0.015%

7.8% + 0.96% + 0.12% + 0.015% = 8.895%