Question

Asked Jan 19, 2019

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The probability that a person in the United States has Type B+ blood is 8%. Four unrelated people in the United States are selected at random. Complete parts (a) through (d).

(a) Find the probability that all four have type B+ blood.

(b) Find the probability none have B+ blood.

(c) Find the probability that one of the four have B+ blood.

(d) Which of the events can be considered unusual? Explain.

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Step 1

Solution: We are given that the probability that a person in the United States has Type B+ blood = 0.08. Also we are told that four unrelated people in the United States are selected at random.

(a) We have to find here the probability that all four have type B+ blood.

Since the events are independent, therefore, we have:

Probability that all four have B+ blood = 0.08 x 0.08x 0.08x0.08

= 0.000041

**Therefore, the probability that all four have type B+ blood is 0.000041.**

Step 2

(b) We have to find the probability that none have B+ blood.

Using the complementary law of probability we have:

Probability that blood type is not B+ = 1 - 0.08

= 0.92

Therefore, the probability that none have B+ blood = 0.92 x 0.92 x 0.92 x 0.92

=0.716393

**Therefore, the probability that none have B+ blood is 0.716393**

Step 3

(c) We have to find the probability that at least one of the four have B+ blood.

The probability that at least one of the four have B+ blood = 1 - Probability that none have B+ blood type

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