1. A question about binomial distributions and conditional probability

1. A family has 2 children. Assuming that the sex of a child is independent of another child, and that male and female children are equally likely, calculate the following probabilities.
a) What is the probability that both children are girls?

This problem follows a binomial distribution, so
P(X=2) = b(2;2, 0.5) = 0.25

b) What is the probability that at least one of the children is female ?
P(X>=1) = b(1;2, 0.5) + b(2;2, 0.5) = 0.5

c) What is the probability that both children are girls if you know that there is at least one daughter?
Essentially the problem is asking for P(a|b), where a and b are the answers for the previous two questions. Since the beginning of the problem states that a and b are independent,
P(a|b) = P(a) = 0.25

The answer doesn't really make intuitive sense to me, so if anyone sees a silly mistake I made somewhere, I'd appreciate you posting about it. Thanks!

2. Re: A question about binomial distributions and conditional probability

Hey RubberDucky.

The second answer is not correct: Using R I got P(X >= 1) = 0.75 and this makes sense since the probability of 0 females is P(X = 0) = 0.25 and 1 - 0.25 = 0.75.

With regards to c) you are right about independence, but you have to remember what the probability and event is here. You are only looking at one child in this case since you know that you have one daughter.

Since P(First Child|Second Child=Daughter) = P(First Child), it means you are looking at probabilities involving only one child which means you have two options: daughter or son and they have equal probability of 0.5.

Hence the final probability of P(First child = daughter|Second child=daughter) = P(First child = daughter) = 0.5

3. Re: A question about binomial distributions and conditional probability

Originally Posted by RubberDucky
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This problem follows a binomial distribution, so
P(X=2) = b(2;2, 0.5) = 0.25
Do you really need to use a formula? The probability the first child is a girl is 1/2. Since the occurrences are independent, the probability the second child is a girl is 1/2. The probability both are girls is (1/2)(1/2)= 1/4.

Another way: writing "G" for girl and "B" for boy, and writing the sex of the first child first, the 2(2)= 4 possibilities are "GG", "GB", "BG", and "BB". Since "Boy" and "Girl" are "equally likely" (probability is 1/2), those four combinations are equally likely. There are 4 of them and 1 is "GG": probability of "GG" is 1/4.

"What is the probability that at least one of the children is female". "At least one" could be "BG" or "GB" or "GG". Each or those, as above, has probability (1/2)(1/2)= 1/4. Since they are "mutually exclusive", the probability is 1/4+ 1/4+ 1/4=3/4, NOT 1/2. Equivalently, of the four "BB", "BG", "GB", "GG" outcomes, three of them give "at least one girl".

"What is the probability that both children are girls if you know that there is at least one daughter." Of the three "BG","GB", and "GG"cases, exactly one corresponds to "both children are girls" is 1/3, NOT 1/4.

(You say "since the beginning of the problem states that a and b are independent". No, it doesn't. It says "Assuming that the sex of a child is independent of another child" which is NOT the same as saying "having at least one daughter" (a) is independent of "having two daughters" (b).)

4. Re: A question about binomial distributions and conditional probability

Okay, thanks to both of you and no, I didn't really need to use the binomial formula now that I've seen you explain an easier way, but my professor isn't really that clear when he teachers (and his accent makes him hard to understand), and when I don't understand concepts in math, I tend to blindly follow formulas XD