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I am trying to solve a probability problem with coins of two types. I will refer to them as circles and squares. The task is:

A boy has 4 circles and 3 squares in his left pocket and 2 circles and 1 square in his right pocket. The boy transfers 2 random objects from his left pocket to his right pocket. Then he transfers two random objects from his right pocket back to his left pocket. The boy then pulls an object from his right pocket. What is the probability he has pulled a circle?

I tried forming Hypotheses $H_{cc}, H_{cr}, H_{rr}, F_{cc}, F_{cr}, F_{rr}$ for the corresponding transfers:

  • Let $H_{ij}$, for $i,j\in\{c,r\}$, represent taking two circles, a circle and a rectangle, or two rectangles in the transfer from the left to the right pocket.

  • The definition of $F_{xy}$ is similar, but in the opposite direction.

I realised it's unwise to then calculate $P(F_{xy}|H_{ij})$, so I gave up on that idea. I considered the following aproach instead:

  • Let $k$ be the number of cirlces moved from the left pocket to the right pocket.

  • Let $l$ be the number of circles moved from the right pocket to the left pocket.

Thus the probability of

  • moving $k$ circles to the right pocket is $\frac{C^k_4\times C^{2-k}_3}{C^2_7}$
  • moving $l$ circles back to the left pocket is: $\frac{C^l_{2+k}\times C^{2-l}_{1+2-k}}{C^2_5}$
  • pulling a circle out of the right pocket after moving two object from left to right and then two object from right to left is $\frac{2+k-l}{3}$

How can I calculate the probabilty?

If $A=\{\text{pulling a circle from the right pocket after the transfers}\}$, then $$P(A)=\sum{P(A|F_{xy})P(F_{xy})}$$
I am not sure what to do next.

EDIT:

If I rename my hypotheses as $H_k$ and $F_l$, where $k$ and $l$ are the number of circles transferred, resp, left $\to$ right and right $\to$ left, then $$P(A)=\sum_{l=0}^2{P(A|F_l)P(F_l)}$$ but $P(F_l)$ should be $P(F_l|H_k)P(H_k)$, therefore $$P(A)=\sum_{k=0}^2{\sum_{l=0}^2{\frac{2+k-l}{3}\times\frac{C^l_{2+k}\times C^{2-l}_{1+2-k}}{C^2_5}\times\frac{C^k_4\times C^{2-k}_3}{C^2_7}}}$$

Am I correct?

BCLC
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Kotaka Danski
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    To simplify this significantly, the second transfer is entirely irrelevant (in the same way that you don't need to care about the top two cards in a deck of playing cards when you're asked about the color of the third one). – Arthur Jan 10 '23 at 11:15
  • @Arthur This can be seen by simplifying the sum expression, I guess. If I remember correctly, the result should depend on the proportions in the first set, right? The answer I get, when I put the sum in Wolfram Alpha is $\frac{22}{35}$ – Kotaka Danski Jan 10 '23 at 11:26
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    I made a miscalcution in my first effort. My answer is repaired now and agrees with the one of @trueblueanil. – drhab Jan 10 '23 at 12:59
  • @drhab I was going to ask for additional explanations but your answer disappeared. Thank you for clearing things up. – Kotaka Danski Jan 10 '23 at 13:14

2 Answers2

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You can simplify it greatly by treating it as a mixture problem.

You are initially having $4C,3S$ in left pocket, $2C,1S$ in right.

By transferring two coins from left to right,
right pocket holds $(2 + \frac47\cdot2) = \frac{22}7 C$ in a total of $5 = \frac{35}7$,

and as @Arthur says in comments, $2_{nd}$ transfer is irrelevant

Thus P(C transferred back) $= \frac{22}{35}$

true blue anil
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Method: for each of the circles calculate the probability to become the coin that is finally selected. You are dealing with $4+2=6$ mutually exclusive events so can find the probability by summation of the $6$ probabilities.

Fix some coin the in left pocket.

The probability that it will become the one eventually selected is $\frac27\frac35\frac13=\frac2{35}$.

Fix some coin in right pocket.

The probability that it will become the one eventually selected is $\frac35\frac13=\frac15$.

So the probability that eventually a circle is selected is:$$4\times\frac2{35}+2\times\frac15=\frac{22}{35}$$

drhab
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