Challenging probability exercise

logicprobabilityprobability theory

I'm dealing with the following exercise (don't be scared about the lenght of the exercise, it has been almost solved):

Mr Smith has 3 boxes of chocolates. The first box contains 20 bars of milk chocolate, 12 bars of dark chocolate and 8 bars of white chocolate. The second box contains 10 bars of milk chocolate, 10 dark chocolates and 10 milk chocolates.
The third box has 10 milk and 10 dark chocolates.

Mr Smith tosses a fair coin twice for choosing from which box he takes a bar of chocolate.

If, by tossing a coin twice, he gets "Head" twice in a row, Mr Smith randomly chooses a bar of chocolate from the first box; if he gets tails twice in a row,Mr Smith randomly chooses a bar of chocolate from the second box; otherwise he randomly chooses from the third box.

It is clear,that once the box has been chosen, the flavour of the chocolate bar is randomly chosen.

(a) What's the probability that the first bar of chocolate is a milk chocolate?

My answer:

I call $A$ the event of choosing the first box. $P(A)=1/4$;

I call $B$ the event of choosing the second box. $P(B)=1/4$;

I call $C$ the event of choosing the second box. $P(B)=1/2$;

And $P(M|A)=1/2,P(M|B)=1/3,P(M|C)=1/2$

Let P(M) the probability of picking milk as first chocolate flavour, and it is equal to:

$P(M)=P(A \cap M)+P(B \cap M)+P(C \cap M)$

And this also reads as:
$P(M)=P(M|A)P(A)+P(M|B)P(B)+P(M|C)P(C)=11/24$

Is that correct?

(b) Mr Smith has drawn a milk chocolate. What's the probability that it comes from the first box?

Applying the Bayes theorem:

$P(A|M)=\frac{P(M|A)P(A)}{P(M)}=\frac{(1/2)(1/4)}{11/24}=3/11$

Is that correct?

(c) Mr Smith has just eaten the first chocolate bar and it was a milk chocolate. What's the probability that the second chocolate bar taken from the same box is again a milk chocolate?

Answer:

I define as $P(M_2)$ the probability that the second chocolate bar is milk chocolate.

Then,

$P(M_2|M)=P(M_2 \cap A|M)+P(M_2 \cap B|M)+P(M_2 \cap C|M)$
which should read as:

$P(M_2|M)=P(M_{2} | A \cap M)P(A|M) + P(M_{2} | B \cap M)P(B|M) + P(M_{2} | C \cap M)P(C|M) $

Here I'm stuck because the professor's slides provide an hint, saying that I should end up with computing the following probability:

$P(M_2|M)=P(M_2|A)P(A|M) + P(M_2|B)P(B|M) + P(M_2|C)P(C|M)$

And this seems to occur iff $P(M_{2} | A \cap M) = P(M_2|A)$, and the same for B,C. Why is that possible? I guess, e.g., $P(M_2|A \cap M)=19/39$. right? Maybe, I can have the equality between $P(M_2|A \cap M)=P(M_2|A)$, if I interpret the event $M_2$ as the second milk chocolate drawn, and conditioning on $M$ is just redundant. Is that correct?

Best Answer

This is a two steps experiment that can be sketched as follows:

                             20/40---- M1
                            /
       --------------(Box 1)--12/40--- D1
      /                     \
     /                       8/40----- W1
  1/4
  /                         10/30---- M2
 /                         /
O--- 1/4 -----------(Box 2)--10/30--- D2
 \                         \
  \                         10/30---- W2
   1/2
     \                       10/20---- M3
      \                     /
       --------------(Box 3)--10/20--- D3

The probability space modelling the question has the final states, the probability of each such final state is the product of the mentioned weights. Now for the questions:

(a) To get a milk (M) chocolate, we must land after the two steps in one of the states M1, M2, or M3. This happens with probability: $$ \frac 14\cdot\frac {20}{40} + \frac 14\cdot\frac {10}{30} + \frac 12\cdot\frac {10}{20} = \frac18+\frac 1{12}+\frac 14=\frac1{24}{3+2+6}=\frac{11}{24}\ . $$

(b) We have the information that Mr. Smith took an M-piece, so the probability space gets restricted to the $3$-elements set containing only M1, M2, and M3:

                             20/40---- M1
                            /
       --------------(Box 1)
      /                     
     /                      
  1/4
  /                         10/30---- M2
 /                         /
O--- 1/4 -----------(Box 2)
 \                         
  \                         
   1/2
     \                       10/20---- M3
      \                     /
       --------------(Box 3)

The total probability for this event M = $\{$ M1 , M2 , M3 $\}$ is computed above, it is $\frac {11}{24}$, so M1 is taken with corresponding probability $$ \frac 14\cdot\frac {20}{40}\ \Big/\ \frac {11}{24} = \frac18\ \Big/\ \frac {11}{24} = \frac 3{11}\ . $$

(c) The question is slightly unclear, so i will make it precise. Mr. Smith took a milk chocolate at the first step. We have thus the space depicted above, in (b), and there is a not mentioned factor of $11/24$ to adjust the probability of each end. So we have the cases:

  3/11---- M1
 /
o--2/11--- M2
 \
  6/11---- M3

Now i assume that we take the "same box", and chose from it randomly. I will denote by M-something the case of a further milky piece, by X-something the other case. Our picture becomes:

                19/39--- M11
               /
    3/11---- M1--20/39-- X1
   /      
  /             9/29---- M22
 /             /
o----2/11--- M2
 \             \
  \             20/29---- X2
   \
    6/11---- M3--9/19---- M33
               \
                10/19---- X3

So the second piece is milky with probability: $$ \frac 3{11}\cdot\frac{19}{39} + \frac 2{11}\cdot\frac{ 9}{29} + \frac 6{11}\cdot\frac{ 9}{19} \ , $$ and the terms should match those from the slides...

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