It is known that the probability that two twins are the same sex $p=0.64$ where in general the probability of the births of a boy is $q= 0.51$. Find the probability that the second of the twins is a boy under the condition that the first of them is a boy.
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There are four possible outcomes $S = \{BB, BG, GB, GG\}$. It is given that
$$ P(BB) + P(GG) = 0.64 $$
It is stated that the probability that the birth of a boy is $0.51$.
$$ P(\text{No Boy}) = 1 - P(\text{Boy}) = 1-0.51 = 0.49 = P(GG)$$
Hence $P(BB) = 0.15$. This gives us $P(BG) + P(GB) = 0.36$ and assuming that $P(BG) = P(GB)$ based on the information provided in the question, we get $P(BG) = 0.18$. To answer the original question,
$$P(\text{Second twin is boy}|\text{First twin is boy}) = \frac{P(BB)}{P(BB)+P(BG)} = \frac{0.15}{0.15+0.18} = \frac{5}{11}$$
There are for combinations boy|boy, boy|girl, girl|boy and girl|girl which we define as $f_{00},f_{01},f_{10},f_{11}$.
We are given that "probability that two twins are the same sex $p=0.64$" which is $f_{00}+f_{11}=0.64$.
We are given that "general the probability of the births of a boy is q=0.51" which is "$f_{00}+f_{01}=0.51$ and $f_{00}+f_{10}=0.51$.
Given that the total probability is $f_{00}+f_{01}+f_{10}+f_{11}=1$.
We invoke our knowledge of algebra, so after removing $f_{00}+f_{11}=0.64$ we have $f_{01}+f_{10}=0.36$ adding $f_{00}+f_{01}=0.51$ we get $2f_{01}+f_{00}+f_{10}=0.87$ and given that $f_{00}+f_{10}=0.51$ we get $f_{01}=0.18$ and $f_{00}=0.51-0.18=0.33$.