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Sickle-cell anemia is an inherited disease in which red

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 31E Chapter 2.3

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 31E

Sickle-cell anemia is an inherited disease in which red blood cells are misshapen and sticky. Sickle cells tend to form clumps in blood vessels, inhibiting the flow of blood. Humans have two genes for sickle-cell anemia, either of which may be S for normal cells or sfor sickle cells. A person with two copies of the s gene will have sickle-cell anemia. A person with one s gene and one S gene will not have the disease, but will be a carrier, which means that the s gene may be transmitted to the person's offspring. If two carriers have a child, the probability is 0.25 that the child will have the disease and 0.5 that the child will be a carrier. Outcomes among children are independent.

a. A mother and father who are both carriers have two children. What is the probability that neither child has the disease?

b. What is the probability that both children are carriers?

c. If neither child has the disease, what is the probability that both are carriers?

d. A woman who is the child of two carriers has a child by a man who is a carrier. What is the probability that this child has the disease?

Step-by-Step Solution:

Solution

Step 1 of 4 :

Given if  two carriers have a child.

First one is a mother carrier and Second one is fother carriers.

Here A be the event.

Let A be the child has the disease and

Here children are independent.

The probability each child disease is 0.25 and

The child will be a carrier is 0.5.

Our goal is to find :

a). Find the probability that neither child has the disease.

b). Find the probability that both children are carriers.

c). Find the probability that both are carriers.

d). Find the probability that this children has the disease.

a).

Now we have to find the probability that neither child has the disease.

1-P(A) = the child first does not have the disease and

1-P(A) = the child second does not have the disease.

P(either child has the disease) = P(the child first does not have the disease and the      

                                                         child first does not have the disease)

P(either child has the disease) = (1- P(A)) (1- P(A))

P(either child has the disease) = (1-0.25) (1-0.25)

P(either child has the disease) = (0.75) (0.75)

P(either child has the disease) = 0.5625

Therefore the probability that neither child has the disease is 0.5625


Step 2 of 4

Chapter 2.3, Problem 31E is Solved
Step 3 of 4

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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