Conditional Probability

Okay here is the question and below it is my attempt to solve it.
1.4 Suppose that the probability that both of a pair of twins are boys is
0.30 and that the probability that they are both girls is 0.26. Given
that the probability of the first child being a boy is 0.52, what is the
probability that the second twin is a boy, given that the first is a boy?



I tried to use bayes theorem as follows.

P(b1)=.52
P(b1|b2)=.3
P(b2)=??
P(b2|b1)=??

P(b2|b1)= (.3*.5)/.52

The correct answer is listed as P(b2|b1)=15/26 =.5777

why did they multiply .52 by .5????

Hello, arascon!

I think you are misreading the given probabilities.

Suppose that the probability that both of a pair of twins are boys is 0.30
and that the probability that they are both girls is 0.26.
And suppose that the probability of the first child being a boy is 0.52,

What is the probability that the second twin is a boy, given that the first is a boy?

Click to expand...

We are given:

. . \(\displaystyle P(\text{both boys}) \:=\:0.30\)

. . \(\displaystyle P(\text{1st boy}) \:=\:0.52\)

\(\displaystyle \text{Bayes' Theorem: }\(\text{2nd boy}\,|\,\text{1st boy}) \;=\;\dfrac{P(\text{both boys})}{P(\text{1st boy})} \;=\; \dfrac{0.30}{0.52} \;=\) .\(\displaystyle \dfrac{30}{52} \;=\;\dfrac{15}{26}\)

Thank you soroban, this is starting to make more sense. Quick question am I wrong in thinking this is Bayes Theorem?



What happened to P(A)? if P(B|A) = .30 and P(B)=.52

Thanks again for your help, I've been trying to get this answer all day. You don't know how much this helps me.