<s>[INST] Conditional probability of cows the question is that if you are a farmer and own six cows: 3 white, 2 black and one that is black on one side and white on the other. Then if you see two black cows (that is 2 black sides of cows) then what is the probability that one of them is the black and white cow?
Here is my attempted answer:
if $M_1$ and $M_2$ are the events that the first or second cow is the mixed one and $b_1$ and $b_2$ denote that the sides of the first and second cow we see is black. then we are looking for $$P(M_1 \cup M_2 \mid b_1b_2)=\frac{P(M_1b_1b_2)+P(M_2b_1b_2)}{P(b_1b_2)}$$
$$=\frac{2P(M_1)P(b_1\mid M_1)P(b_2\mid M_1b_1)}{P(M_1 \cup M_2)P(b_1b_2\mid M_1 \cup M_2)+[1-P(M_1 \cup M_2)]P(b_1b_2\mid M_1^cM_2^c)}$$
then $P(M_1)$ = $\frac{1}{6}$ as there are $6$ sheep and only $1$ that is mixed. $P(b_1\mid M_1)= \frac{1}{2}$ as it can be one of two sides and $P(b_2\mid M_1b_1)=\frac{2}{5}$ as this is just the probability of choosing a black cow. So the numerator is equal to $\frac{1}{15}$.
$P(M_1 \cup M_2) = P(M_1) + P(M_2)$ as they are disjoint and so is equal to $\frac{1}{3}$. Then $P(b_1b_2\mid M_1 \cup M_2)$ is just the probability that the other cow is black and we see the black side of the mixed sheep and so is $\frac{2}{5} *\frac{1}{2}$. Finally $[1-P(M_1 \cup M_2)]=\frac{2}{3}$ and $P(b_1b_2\mid M_1^cM_2^c)$ is $\frac{1}{10}$ as it is the number of all black pairs over the total number of pairs. so putting it all together. I get $\frac{\frac{1}{15}}{\frac{2}{15}}=\frac{1}{2}$ but the answer is supposedly $\approx .3$ 
EDIT
The guy who wrote the paper made a mistake in the answers. It should be $\frac{1}{2}$
 [/INST] If the question is asking for the probability that either of the two cows is 2-coloured, we have
$$P(\text {1 cow is 2-coloured | both visible sides are black}) = \frac{P(\text {1 cow is 2-coloured and other is black}) \times P(\text {the black side of the 2-coloured cow is seen})}{P(\text{both visible sides are black})}=\frac{\frac{\binom{3}{0}\binom{1}{1}\binom{2}{1}\cdot\frac{1}{2}}{\binom{6}{2}}}{\frac{\binom{3}{0}\binom{1}{1}\binom{2}{1}\cdot\frac{1}{2}}{\binom{6}{2}}+\frac{\binom{3}{0}\binom{1}{0}\binom{2}{2}}{\binom{6}{2}}}=\frac{1}{2}$$
where $$\frac{1}{15}=\frac{\binom{3}{0}\binom{1}{1}\binom{2}{1}\cdot\frac{1}{2}}{\binom{6}{2}}$$ is the probability that the $2$ visible sides are black when one is 2-coloured and the other is black and $$\frac{1}{15}=\frac{\binom{3}{0}\binom{1}{0}\binom{2}{2}}{\binom{6}{2}}$$ is the probability that the $2$ visible sides are black when both cows are black (these exhaust all possibilities for both visible sides being black).
If the question is asking for the probability that a certain (of the two) cow be 2-coloured, the probability is then $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$
Note that the number of white cows is irrelevant.
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