An exercise of probability

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.

  1. 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.

  1. 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.

  1. 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?

Best Answer

Let $R_mG_n$ represent the event of choosing $m$ red and $n$ green apples. We now have

  1. 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 $$P_1=\frac23\left(P(R_2G_1)+P(R_1G_2)\right)=\frac23\cdot\frac{2\cdot\binom52\binom51}{\binom{10}3}=\boxed{\frac59}$$

  2. 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,

$$P_2=\frac13\left(P(R_2G_1)+P(R_1G_2)\right)+P(R_3G_0)+P(R_0G_3)$$ $$P_2=\frac13\cdot\frac{2\cdot\binom52\binom51}{\binom{10}3}+\frac{2\cdot\binom53}{\binom{10}3}=\boxed{\frac49}$$

  1. 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.

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