Probability Problem Having to Do With Cystic Fibrosis?
Monday, October 4th, 2010 at
8:54 pm
Aurore asked:
It is estimated that 1 in 25 Americans is a cystic fibrosis carrier. Find the probability that a randomly selected American couple’s child will have cystic fibrosis, assuming that they’re unrelated?
How do I do this? Thanks for your help; I want to understand the process.
Tagged with: Cystic Fibrosis • S Child • Thanks For Your Help
Filed under: Cystic Fibrosis
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probability of father a carrier is 1/25
probability of mother a carrier is 1/25
probability of both is 1/25 * 1/25 = 1/625 which is th probability of a cf baby.
I believe the CF gene is recessive. So if two parents are carriers, the child needs to get the recessive gene from each parent to suffer from CF. In other words, if both parents are carriers, it’s a 1/4 probability (1/2 from the father x 1/2 from the mother). (And I know a family where only one child has CF).
probability father is a carrier is 1/25
probability mother is a carrier is 1/25
So to get CF, all 3 events must happen (father is a carrier, mother is a carrier, gets both recessive genes), so it’s 1/25 x 1/25 x 1/4 = 1 / 2500