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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 $\approx .3$


The guy who wrote the paper made a mistake in the answers. It should be $\frac{1}{2}$

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Natural language questions about probability can often lead to many interpretations. A similar problem is the boy born on a Tuesday problem which also appeared on MSE before. – Willie Wong Nov 28 '12 at 15:40
Given your priors (each cow is equally likely to appear in front of you, each cow is equally likely to show you either of its two sides), your Baysian analysis is okay. As Johan indicated, "one of them" may mean a specific one. In which case you have to divide your probability by 2 to get 1/4 = 0.25 which some people would round up to be approximate 0.3. But with all Bayesian analysis you have to be careful to state your priors. – Willie Wong Nov 28 '12 at 16:23
For example, as a farmer I may notice that my cattle tends to face one way only and so if I know which side of my two-coloured cow is black and which side is white (I better, if I am the farmer), the probability is either 0 (if north is such that my bicolour cow would always show me her white side) or 2/3 or 1/3 (if north is the other way, and depending on whether Johan's interpretation is taken). – Willie Wong Nov 28 '12 at 16:25
up vote 1 down vote accepted

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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Why is the probability of 1 cow being 2 coloured and the other black equal to 1/18? If each cow is equiprobable, the probability should be $2/15$ as Ben calculated in the OP. – Willie Wong Nov 28 '12 at 15:32
@WillieWong Thanks for pointing out the error; I have fixed it. – Sarastro Nov 29 '12 at 11:45

This problem can be solved by listing all the possibilities.

If you see two black cows, you could be seeing:

  • Black cow 1, Black cow 2
  • Black cow 1, Two-side cow
  • Black cow 2, Two-sided cow

So, if you see two black cows, 2 out of 3 times one of them will be the two-sided cow. Therefore, the probability is 0.66.


I think Johan is right. I interpreted the question as

If you see two black cows, what is the probability that either of the cows is the white/black cow?

I think the answer of 0.3 would be correct if you interpreted the question like this:

If you see two black cows, what is the probability of selecting the two-side cow if you pick one of them at random?

In that case the answer would be the probability that one of the cows is the two-sided one (0.66) times the probability that you pick the two-sided cow assuming it's there (0.5). That yields a probability of about 0.3.

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But doesn't this give equal weight to them all when in fact it should be less weight given to the double sided ones as they are less likely to happen. Also the answer is approximately $.3$ – Ben Nov 28 '12 at 13:02
Your question reads "if you see two black cows". Did you mean "if you see two cows"? – woz Nov 28 '12 at 13:04
no its if you see two black cows. But by seeing them do you not add weight to the possibility that they are both black – Ben Nov 28 '12 at 13:06
If you give the condition "if you see two black cows", then you assume exactly that. The probability of seeing a white cow is zero if you start with the condition "if you see two black cows". – woz Nov 28 '12 at 13:09
The calculation by woz is correct. The only thing I can suggest is that when you ask "what is the probability that one of them is the black and white cow" you mean a specific one. (Say the one on the left.) That probability is approximately $0.33$ since it happens in exactly half of the cases woz computed the probability for. – Johan Nov 28 '12 at 14:15

Let $B$ be the event that we see two black sides.

Since one can choose two cows in ${6\choose 2}=15$ ways the probability that the two black cows are chosen is ${1\over15}$, and the probability that the black-white cow and a black cow is chosen is ${2\over15}$. In the latter case the probability that we actually see two black sides is only half of that.

Since there are no other ways that we can see two black sides we have $$P(B)={1\over 15}+{1\over2}\cdot{2\over15}\ .$$

Now exactly half of this probability comes from sightings of two black cows and the other half from sightings of the black-white and a black cow. Therefore the conditional probability of one cow being black-white when we see two black sides is ${1\over2}$.

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