One out of 25 healthy people carries a single gene for cystic fibrosis (CF), these people...

One out of 25 healthy people carries a single gene for cystic fibrosis (CF), these people are called carriers and healthy people without a CF gene are called non-carriers. A uniformly-chosen random healthy person has probability 1/25 of being a carrier.
A person with two CF genes is not healthy; they are sick (with cystic fibrosis). The child of a carrier has probability 1/2 of inheriting a CF gene from that parent. The child of two carriers inherits each of their parents CF genes independently, so the child has probability 1/4 of having CF.

Two uniformly-chosen random healthy people have a child together. What is the probability that the child has CF?
Two uniformly-chosen random healthy people have a child together. What is the probability that the child is a healthy non-carrier?
A carrier has a child with a uniformly-chosen random healthy person. What is the probability that the child has CF?
A carrier has a child with a uniformly-chosen random healthy person. What is the probability that the child is a (healthy) carrier?
Two uniformly-chosen healthy people have a baby. A quick blood test, administered minutes after birth, shows that the baby has at least one CF gene, but gives no other information. What is the probability that the baby has CF?

Ok so this is how I solved it (I have not included all my steps as they are long):

for number 1, I figured the probability is 1/2500 because the probability of a child having CF is = to the probability of the father being a carrier * by the probability that the mother is a carrier * by the probability that it inherits both CF genes... 1/25 * 1/25 * 1/4...

for number 2, I got 2401/2500 because pr of child is a healthy non carrier is = (pr of father being a carrier * pr of mother being a non carrier)
+ (pr of father is non carrier * pr mother being a carrier * pr of inheriting no CF gene)
+ (pr of father is non carrier * mother being a non carrier)
+ (pr of father is carrier * pr of mother is carrier * inheriting no CF gene)
... calculating these probabilities gave me the answer above.

for number 3, I got 1/25 * 1/4 because the probability of it inheriting both genes is 1/4 * by the probability of the other parent being a carrier is 1/25 ... I got 1/100

for number 4, I got 49/1250 (this calculation was similar to #2 but a bit longer)

for number 5, here we have the probability of a baby having CF given that the baby has one CF gene:
P(baby has CF) = 1 - P(healthy non carrier) - P(healthy carrier)
= 1 - 2401/2500 - 98/2500
= 1 / 2500
and the P(has CF given at least one CF gene)
= 1/ 2500
-------------
1/2500 + 98/2500
=1/99
Please let me know if I computed these correctly, if not to try and guide me towards the direction!
thanks!

Solutions

Expert Solution

First is correct.

Second is also correct including the logic and the computation.

In the third part, you have to multiply it by 2 since the given carrier could be mother/father. Hence two possible cases and the same computation. Your answer should be 1/50.

In the fourth part, it is given that one partner is a carrier. Hence possible cases are -

(father being a carrier * pr of a mother being a non-carrier * pr of inheriting CF gene) +
(pr of a father is non-carrier * mother being a carrier * pr of inheriting CF gene) + (pr of father is carrier * pr of mother is carrier * pr of inheriting one CF gene) = (1* 24/25*1/2) + (24/25*1 1/2) + (1/25*1/2) = 49/50

[When both are a carrier then the child can have one carrier = Inherit from father*Do not Inherit from mother + Do not Inherit from father*Inherit from mother = 1/2*1/2 + 1/2*1/2 = 1/2]

In the fifth part, Pr(baby has at least one CF gene) = 1 - Pr(baby has no CF gene) = 1 - (24/25*24/25) = 49/625

Pr(baby has CF gene | baby has at least one CF gene) = Pr(baby has CF gene and the baby has at least one CF gene) / Pr(baby has at least one CF gene) = Pr(baby has CF gene)/Pr(baby has at least one CF gene) = (1/2500)/(49/625) = 1/196

orchestra answered 1 year ago

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