Genetics and statistics

[MedGrl@2022]

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205. Colorblindness is a sex-linked recessive trait. If a colorblind man and a woman that is a carrier for the trait have two girls and two boys, what is the probability that at least one of the girls will be colorblind?
A. 0%
B. 50%
C. 75%
D. 100%

I choose B (even though I originally thought that the answer was 25%, but that was not an option) but the answer is C.

This is Examkrackers explanation: “C is correct. There is a 50% probability for each girl to be colorblind. This means that there is a 25% probability that both will be colorblind and 25% that neither will be colorblind, leaving 50% that either one or the other might. The boys are irrelevant. We want all the situations where at least one is colorblind. 50 + 25 = 75.”

I understand that “there is a 50% probability for each girl to be colorblind. This means that there is a 25% probability that both will be colorblind and 25% that neither will be colorblind”. What is EK saying about “leaving 50% that either one or the other might” and then how did they figure adding the two numbers together? I think it summing the probabilities but I am confused.

Is the first 50 for one girl having the disorder and the 25 for both of them having the disorder? Why isn’t it just 50%?

Thank you,

Verónica

 

[milski]

Yes, at least one of the girls includes also the possibility of both of the girls being affected. The four possible outcomes are SS, SH, HS and HH. Of these, 3 have at least one of the girls sick, so the final result is 3/4=75%.

If the probability of something happening is rather messy, it might be easier to calculate the probability of the event not happening and know that the probability you're looking for is 1 minus that.

For example, what is the probability of two people having different birthdays? It will take a while to construct all the possible cases. On the other hand it's easy to see that the probability of them having the same birthday is 1/365. Thus, the probability of having different birthdays is 1-1/365=364/365.

 

[MedGrl@2022]

Yes, at least one of the girls includes also the possibility of both of the girls being affected. The four possible outcomes are SS, SH, HS and HH. Of these, 3 have at least one of the girls sick, so the final result is 3/4=75%.

If the probability of something happening is rather messy, it might be easier to calculate the probability of the event not happening and know that the probability you're looking for is 1 minus that.

For example, what is the probability of two people having different birthdays? It will take a while to construct all the possible cases. On the other hand it's easy to see that the probability of them having the same birthday is 1/365. Thus, the probability of having different birthdays is 1-1/365=364/365.

Wow thanks I didn't think of that... it is so much easier to look at it that way! 🙂