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probability and genetics question


Full Member
Jun 15, 2020
  1. Pre-Medical
Suppose the brown allele for eye color (B) is completely dominant over the blue allele for eye color (b). If two brown-eyed parents produce a child that is blue-eyed, what is the probability that at least one out of the next two children they produce will also have blue eyes?Choose 1 answer:

Choose 1 answer:

(Choice C)C44%

I understand how to get to the correct answer, but my question is: why can I not use the rule of addition to just add probability that the first child has blue eyes (1/4) and the probability the second one does...so, 1/4 + 1/4 = 50%

i.e.. For mutually exclusive events X and Y, the probability (P) that one will occur (X or Y) is P(X)+P(Y)


New Member
2+ Year Member
Mar 15, 2017
  1. Medical Student (Accepted)
You want at least one of the kids to have blue eyes, so you have three options: either the 1st one is blue eyed, the 2nd is blue eyed, or both are blue eyed so you add the three probabilities.

If only one is blue eyed, then the other is brown eyed so what is the probability of having a blue eyed kid and the other one being brown eyed ? It's 1/4 x 3/4 = 3/16
What is the probability of having one kid blue eyed and the other one blue eyed? It's 1/4 x 1/4 = 1/16
Now add the probabilities : 3/16 + 3/16 + 1/16 = 7/16

Alternatively you can deduct the complement from 1.
The complement of AT LEAST events is NONE i.e. what is the probability of neither kid being blue eyed = both being brown eyed? It's 3/4 x 3/4 = 9/16
The probability of at least one being blue eyed is therefore 1 - 9/16 = 7/16

This explanation helped me understand, see if it does the same for you.

Maybe the correct way to use the rule of addition in this scenario is P(A or B or C) = P(A)+P(B)+P(C), as there are three discrete outcomes. A is kid 2 having blue eyes, while kid 3 has brown. B is the opposite. C is both having blue eyes, at least as far as I understand it. That might be the wrong way of thinking about it, however.
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