Solved – Probability of having 2 girls and probability of having at least one girl

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In this question the general formula P(A|B) = P(A ∩ B)/P(B) is used. I understand through intuition why the answer should be 1/3. What I don't understand is why P(both girls, at least one girl) is 1/4. I know that P(A ∩ B) = P(A)P(B) if A and B are independent. Here P(Both girls) = 1/4 and P(at least one girl) = 3/4 so why isn't P(Both girls, at least one girl) = 3/16?

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Like you said, $P(A \cap B) = P(A) P(B)$ only when $A$ and $B$ are independent. In this instance if $A = $ Both girls, and $B$ = at least one girl, these events are not independent. More specifically, if $A$ holds, then both are girls, which automatically implies that there is at least one girl, so $B$ holds. Thus, you don't have independence.

From the above logic it is thus clear that, if $A$ holds, then $B$ holds, which means that the event $A \subset B$. Thus $A \cap B = A$. So $P(A \cap B) = P(A) = 1/4$ in this case.

Related Solutions

Probability – Understanding Conditional and Unconditional Learning Errors

For (a), a simple way to look at is that you've reduced your probability space to only the combinations that have at least one boy:

BB = 1/3
BG = 1/3
GB = 1/3

GG is no longer a possibility based on the fact that your neighbor said he had at least one boy. Of the possibilities remaining, you're left with a 2/3 probability that he has a girl. The information he gave you reduced the probability of him having a girl from 3/4 to 2/3. Formally, this can be shown as follows: $$P(At\ least\ one\ girl|At\ least\ one\ boy) = \frac{P(At\ least\ one\ girl\ \cap At\ least\ one\ boy)}{P(At\ least\ one\ boy)} $$ From your original box, we can see the probability of having at least one boy and at least one girl is BG + GB = 0.25 + 0.25 = 0.5, but we need to divide by the probability of at least one boy, which is BB + BG + GB = 0.25 + 0.25 + 0.25 = 0.75, so we get $\frac{\tfrac{1}{2}}{\tfrac{3}{4}} = \frac{2}{3}$.

For (b), now that we've seen a boy, the only uncertainty remaining is the gender of the other child, and given no other information, the probability of the other child being female is 1/2, which is the answer.

Solved – the probability that a girl is taller than a boy, with boys ~N(68, 4.5) and girls ~N(62, 3.2)

Assume that $X$ is a random height of a boy and $Y$ is a random height of a girl and these variables are independent. Now, the probability you want to assess is $$ P(X < Y) = P(X - Y < 0). $$ Since $X$ and $Y$ are independent gaussian variables we use a theorem saying that a sum of independent gaussians is gaussian with a mean equal to sum of the means and the variance equal to sum of the variances. So $Z = X - Y$ has the distribution with parameters which you calculated.

So you found the parameters and, as long as you have the normalcdf function for any gaussian distribution, you are done. You may be also asked for solution when you are given just a standard gaussian table. Then you need to still simplify.

We are looking for $P(Z < 0)$. Let's say $Z \sim N(\mu,\sigma^2)$. Then $$ P(Z < 0) = P(Z - \mu < -\mu) = P\left(\frac{Z - \mu}{\sigma} < -\frac{\mu}{\sigma}\right). $$ Note that I transformed the inequality such that $$G = \frac{Z - \mu}{\sigma}$$ is standard gaussian. So we look for the probability $$ P\left(G < -\frac{\mu}{\sigma}\right), $$ where the left side is standard gaussian so we can find the result in a table.

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Related Solutions

Probability – Understanding Conditional and Unconditional Learning Errors

Solved – the probability that a girl is taller than a boy, with boys ~N(68, 4.5) and girls ~N(62, 3.2)

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