Question

About 1/4 of all adults in the United States have type O+ blood. If three randomly selected adults donate blood, find the probability of each of the following events. (Round your answers to four decimal places.) (a) All three are type O+. 0.0156 Correct: Your answer is correct. (b) None of them is type O+. 0.4219 Correct: Your answer is correct. (c) Two out of the three are type O+.

Answer 1

Answer:

a) 0.0156

b) 0.4219

c) 0.1406

Step-by-step explanation:

We are given the following information:

We treat adult having type O+ blood as a success.

P(Adult have type O+ blood) = 0.25

Then the number of adults follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3

a) All three are type O+

$P(x =3)\\= \binom{3}{3}(0.25)^3(1-0.25)^0\\= 0.0156$

b) None of them is type O+

$P(x =0)\\= \binom{3}{0}(0.25)^0(1-0.25)^3\\= 0.4219$

c) Two out of the three are type O

$P(x =2)\\= \binom{3}{2}(0.25)^2(1-0.25)^1\\= 0.1406$

Answer 2

Using the principle of binomial probability, the probability o the required events are :

• All three O positive = 0.0156

• None O positive = 0.4219

• 2 O positive = 0.1406

Recall :

• P(x = x) = nCx * p^x * q^(n-x)

• p = probability of success = 1/4 = 0.25

• q = 1 - p = 1 - 0.25 = 0.75

• n = number of trials = 3

A.) Probability that all 3 are O+ :

x = 3

P(x = 3) = 3C3 × 0.25³ × 1 = 0.0156

B.) Probability that none 3 are O+

x = 0

P(x = 3) = 3C0 × 1 × 0.75³= 0.4219

C.) Probability that 2 out of the three are O+

x = 2

P(x = 2) = 3C2 × 0.25² × 0.75¹ = 0.1406

Therefore, the required probabilities are : 0.0156, 0.4219 and 0.1406 respectively.

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