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Jack receives monthly allowance of \$10, \$50 or \$100 with equal probability. Jack puts part of the money into his savings box. Each month he puts \$1, \$5 or \$10 into the box with equal probability. If he receives \$10 he doesn't put \$10 into the box. The probability that Jack will save tenth of the allowance is 2/3. Let $X$ be a variable that counts the amount of money Jack puts into the box in a given month. Let $Y$ be a variable that counts the amount of money Jack receives with allowance for the same month. Find the probability function of the 2 variables.

The $\mathbf{answer}$ is this: enter image description here

However I don't understand how we arrive to this. It's usually nice to solve this problems with a $\mathbf{tree}$ of probabilities so I drew one:

enter image description here For example, $P(X=1, Y = 1) = 1/3 * 1/2 = 1/6$ which is wrong. Also I'm not sure what to do with the fact that the probability that Jack will save tenth of what he received is 2/3.

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    Given the rules of the game, how do you arrive at the statement "The probability that Jack will save a tenth of the allowance is 2/3" ? – 2017-02-25
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    This was given in the problem – 2017-02-25

1 Answers 1

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The trouble lies in the wording of the problem, which is only clarified after seeing the given answer.

Jack receives monthly allowance of \$10, \$50 or \$100 with equal probability.

Straight forward; the marginal distribution of the allowance is uniform over those values.$$\mathsf P(Y{=}k) ~=~ \tfrac 13\quad\Big[ k\in\{10,50,100\}\Big]$$

Jack puts part of the money into his savings box. Each month he puts \$1, \$5 or \$10 into the box with equal probability.

This is the marginal distribution of the savings. $$\mathsf P(X=k) ~=~ \tfrac 13\quad\Big[ k\in\{1,5,10\}\Big]$$

If he receives \$10 he doesn't put \$10 into the box. The probability that Jack will save tenth of the allowance is 2/3.

This looks contradictory. However, its apparent this was intended to convey conditional probabilities.$$\begin{align}\mathsf P(X=10\mid Y=10) ~&=~ 0\\ \mathsf P(X=k\mid Y=10k) ~&=~ \tfrac 23 \quad\Big[k\in\{1,5,10\}\Big]\end{align}$$

and hence $$\begin{align}\mathsf P(X=10, Y=10) ~&=~ \mathsf P(X=10\mid Y=10)\mathsf P(Y=10)~&=&~ 0\\ \mathsf P(X=k, Y=10k) ~&=~ \mathsf P(X=k\mid Y=10k)\mathsf P(Y=10k) ~&=&~ \tfrac 29 \quad\Big[k\in\{1,5,10\}\Big]\end{align}$$

The suggested solution is to construct a joint probability table and use the values for the six margins and four cells to fill in the rest.

$$\begin{array}{c|c:c:c|c} Y\backslash X & 1 & 5 & 10 & \\\hline 10 & 2/9 & & 0 & 1/3 \\ \hdashline 50 & & 2/9 & & 1/3 \\ \hdashline 100 & & &2/9 & 1/3\\\hline & 1/3 & 1/3 & 1/3 & 1 \end{array}$$

For instance knowing that $\mathsf P(X=1,Y=10)=\tfrac 29$ and $\mathsf P(X=10,Y=10)=0$ tells us that $\mathsf P(X=5,Y=10)=\tfrac 19$ because those three must sum to $\mathsf P(Y=10)$ which is $\tfrac 13$.

You can use a tree diagram to arrive at a solution. However, you are not given the second branches; these are what you need to find. You are given the first branches, the final values, and some of the second branch:

$$\begin{array}{l} &&&&\bullet \\ && \swarrow{\tfrac 13} &&\downarrow{\tfrac 13} && {\tfrac 13}\searrow \\ & \circ &&&\circ&&&\circ \\ \swarrow{\tfrac 23} & \downarrow{?} & {0}\searrow & \swarrow{?} & \downarrow{\tfrac 23} & {?}\searrow &\swarrow{?} & \downarrow{?} & {\tfrac 23}\searrow \\ {(\tfrac 29)\\X=1\\Y=10}& {(?)\\X=5\\Y=10}&{(0)\\X=10\\Y=10}&{(?)\\X=1\\Y=50}&{(\tfrac 29)\\X=5\\Y=50}&{(?)\\X=10\\Y=50}&{(?)\\X=1\\Y=100}&{(?)\\X=5\\Y=100}&{(\tfrac 29)\\X=10\\Y=100} \end{array}$$

Which is no where near as convenient for seeing the solution; although it is possible.

$\Box$

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    Thanks you! I agree the wording of the problem seems to be intentionally confusing. Also in your tree diagram I think there's a misspelling with $2/3$ in the right-most branch at the top – 2017-02-25