Show solution and explanation for each. Genetic testing reveals that two newlywe

Question

Show solution and explanation for each.
Genetic testing reveals that two newlyweds are both carriers of the recessive mutation that causes cystic fibrosis. As a consequence, any child produced by the marriage has a probability of Vaof being afflicted with cystic fibrosis. If this couple has three children (none of whom are identical siblings), what is the probability that: (a) none of the children will have cystic fibrosis? (b) at least one child will have cystic fibrosis? (c) one child will have cystic fibrosis and the other two will not? (d) the last-born child will have cystic fibrosis, given that you already know that the first two do not have cystic fibrosis? (show full solution)

Explanation / Answer

Solution;

A) The probability that none of the children will have cystic fibrosis is 0.421875.

Number of total children = 3

probability of gettting cystic fibrosis = 0.25

Number of success = 0

by appplying binomail distribution ;

P(x, n, p) = nCx * px * (1-p)(n-x)

P(x = 0) = 0.421875

B) The probability that at least child will have cystic fibrosis is 0.578125.

Number of total children = 3

Probability of getting cystic = 0.25

Number of success = 1

By applying Binomial distribution:-

P(x, n, p) = nCx * px * (1-p)(n-x)

P(x > 1) = 0.578125

C) The probability that one child will have cystic fibrosis and the other two will not is 0.421875

Number of children = 3

Probability of getting cystic = 0.25

Number of success = 1

By applying Binomial distribution:-

P(x, n, p) = nCx * px * (1-p)(n-x)

P(x = 1) = 0.421875

D) The last born child will have cystic fibrosis, given that you already know that the first two do not have cystic fibrosis is 0.25.

Probability of getting Cystic is independent of other two outcomes so the probability of getting cystic for last born child would be 0.25

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