An exercise of probability Alice has 5 red apples and 5 green apples. She chooses 3 apples at random and puts them in a bag. Alice's dog finds the bag and eats one apple (no matter on which colour it is).
Note that there are $\binom{10}{3}$ possible way of choosing 3 out of the 10 apples.

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*What is the probability that the two remaining apples are of different colors?

My answer: I note that Alice must take 1 red apple, 1 green apple and another apple (no matter on its colour) which has been eaten by her dog. To this regard, she has $\binom{5}{1} \cdot \binom{5}{1} \cdot \binom{8}{1}=200$ ways of choosing the three apples. Thus the probability of choosing such a combination of 3 apples equals to $\frac{200}{\binom{10}{3}}$. At this point, I believe that a conditional probability solves my problem, but I have no clue of how to continue.


*What is the probability that the two remaining apples are of the same color?

My answer: I argue in a very similar way to the first part, and have the same troubles.


*Given that the remaining apples are both red, what is the probability that Alice put in the bag 3 red apples?

I have no clue. I believe that conditional probability and Bayes' theorem will be helpful, but I'm not able to use them. Can you help me, please?
 A: Let $R_mG_n$ represent the event of choosing $m$ red and $n$ green apples.
We now have

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*For the remaining two apples to be differently coloured, we must have chosen $2$ apples of one colour and $1$ of the other. This can be done in $2\cdot\binom52\binom51$ ways. Her dog eats one of the two similarly coloured apples with a chance of $\frac23$ which yields us with a final probability of $$P_1=\frac23\left(P(R_2G_1)+P(R_1G_2)\right)=\frac23\cdot\frac{2\cdot\binom52\binom51}{\binom{10}3}=\boxed{\frac59}$$


*For the remaining apples to be similarly coloured, we can either choose all $3$ of one colour($2\cdot\binom53$ ways) and have the dog eat any of them, or $2$ of one colour($2\cdot\binom52\binom51$ ways) and have the dog eat the oddly coloured one(a $\frac13$ chance). Thus,
$$P_2=\frac13\left(P(R_2G_1)+P(R_1G_2)\right)+P(R_3G_0)+P(R_0G_3)$$
$$P_2=\frac13\cdot\frac{2\cdot\binom52\binom51}{\binom{10}3}+\frac{2\cdot\binom53}{\binom{10}3}=\boxed{\frac49}$$


*The probability of both remaining apples being red is $\frac13P(R_2G_1)+P(R_3G_0)=\frac{P_2}2=\frac29$. Thus,

$$P(R_3|\text{2 red remaining)}=\frac{P(R_3G_0)}{\frac13P(R_2G_1)+P(R_3G_0)}=\boxed{\frac38}$$

Side note: $P_2$ can also be calculated as $(1-P_1)$ because we can only end up with $2$ similarly coloured apples or $2$ differently coloured apples.
