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Suppose that $3$ balls are chosen with replacement from an urn consisting of $5$ white and $8$ red balls. Suppose that the white balls are numbered, and let $Y_i$ equal $1$ if the $i$-th white ball is selected and $0$ otherwise. Find the joint probability mass function of $Y_1,Y_2$.

My attempt is: let $Y_1$ be the event that the white ball number $1$ is chosen and $Y_2$ be the event that the white ball number $2$ is chosen

$$p(0,0)=11*11*11/13*13*13=1331/2197$$

$$p(1,0)=1*11*11/13*13*13=121/2197$$

$$p(0,1)=1*11*11/13*13*13=121/2197$$

$$p(1,1)=1*1*11/13*13*13=11/2197$$

but the sum is not $1$. could someone help me to understand my mistake?

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    But $13^3=2197$. – 2017-01-03
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    Also: You don't consider the placements for the two special balls. In your formula for $p(1,0)$ for example, you assume that $W_1$ is chosen first. – 2017-01-03
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    Also: you exclude cases where, say, $W_1$ is chosen three times. – 2017-01-03
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    @lulu is right that you haven't taken ordering into account, you also haven't taken into account drawing a target ball more than once. – 2017-01-03
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    In that case I choose the white ball number 1 but not the number 2. – 2017-01-03
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    in p(0,0) the white ball number 1 and the white ball number2 are not chosen in p(1,0) the white ball number 1 is chosen and the white ball number2 is never chosen p(0,1) the contrary p(1,1) the two balls are taken, is it wrong? – 2017-01-03
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    is W1 the event that the first ball is chosen? – 2017-01-03
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    Note; if you want to address a comment to a user, you need to put @Anne or the like in it somewhere. $W_1$ is just the first white ball. For $p(1,0)$ you might draw $W_1$ three times, yes? Or two times. You don't consider this case. – 2017-01-03
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    So, for $P(1,0)$ : To draw $W_1$ exactly once is $3\times 11^2 \times \frac {1}{13^3}$. To draw $W_1$ exactly twice is $3\times 11 \times \frac 1{13^3}$. To draw $W_1$ exactly three times is $\frac 1 {13^3}$. Same for $P(0,1)$. Can you handle $P(1,1)$? – 2017-01-03
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    @lulu $$P(1,0)=3*11^2*1/13^3+ 3*11*1/13^3+1/13^3=4753/2197>1?$$ $$P(1,1)=3*11/13^3+ 3*11/13^3=66/2197$$ – 2017-01-03
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    Check your math. Also, don't forget to put @username in your comment or the intended recipient won't get it. – 2017-01-03
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    For $P(1,1)$ you are again neglecting cases like $W_1W_1W_2$. – 2017-01-03
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    $$W1W2W2$$ and the other combinations---->$$3*11/13^3$$ $$W1W1W2$$ and the other combinations---->$$3*11/13^3$$ – 2017-01-03
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    You lost me. What is your answer for $P(1,1)$? – 2017-01-03
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    yes, i lost you, the one in which i'm neglecting the case W1W1W2 – 2017-01-03
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    Well, I'm not understanding what your final answer for $P(1,1)$ is. Can you just state it? – 2017-01-03
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    it seems to me that I didn't take ordering into account so I'm so sorry but I didn't understand your comments Could you explain in other words, please? – 2017-01-03
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    @lulu $$P(1,1)=3*11/13^3+3*11/13^3$$ but why P(0,1)=4753/2197>1? – 2017-01-03
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    Not your username....the name of the intended recipient. That answer for $P(1,1)$ is wrong for the reason I said. You fail to count permutations of $W_1W_1W_2$ and $W_2W_2W_1$. – 2017-01-03
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    I didn't understand , If have 1 time W1 e two times W2 I can have W1W2W2, W2W1W2, W2W2W1, If have 2 times W1 e 1 time W2 I can have W2W1W1, W1W2W1, W1W1W2 in p(1,1) – 2017-01-03
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    Ok...so, again, what is your answer for $P(1,1)$? Also, in computing $P(0,1)$ you appear to simply be adding incorrectly. I should advise: take greater care with your calculations. If you constantly make careless errors, you'll never manage these sort of calculations (which are, admittedly, error prone). – 2017-01-03
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    i'm more and more confused...is it correct my interpretation of your P(1,0) in my post? $$P(1,0)=3*11^2/13^3+3*11/13^3+1/13^3$$ if yes why is it >1? I'm trying to understand your way of solving the problem – 2017-01-03
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    For heaven's sakes. As you have written, $P(1,0)=\frac {3\times 11^2+3\times 11 +1}{13^3}$, yes? Just add the three terms in the numerator. – 2017-01-03
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    could you write your answers? – 2017-01-03
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    @lulu you are not clear at all – 2017-01-03

1 Answers 1

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Let $W_1$ denote the first white ball, $W_2$ the second. We remark that there are $11$ balls aside from those two, thus the probability of choosing a neutral ball is $\frac {11}{13}$. We will let $X$ denote a generic "neutral" choice.

$P(0,0)$: We need each choice to be neutral so $$P(0,0)=\left( \frac {11}{13}\right)^3=\frac {1331}{2197}$$

$P(1,0)$: here we have to distinguish a couple of cases. Case I: you choose $W_1$ exactly once. Then there are $3$ places to locate the special ball, so $\frac {3\times 11^2}{13^3}$.

Case II: you choose $W_1$ exactly twice. Then there are $3$ places to locate the neutral ball so $\frac {3\times 11}{13^3}$.

Case III: You choose $W_1$ three times. $\frac 1{13^3}$

Thus $$P(1,0)=\frac {3\times 11^2+3\times 11+1}{13^3}=\frac {397}{2197}$$

$P(0,1)=P(1,0)$ by symmetry.

$P(1,1)$: you have to distinguish two cases.

Case I: the third ball is neutral. Then there are $3$ ways to place $W_1$ and then $2$ ways to place $W_2$ and then $11$ choices for $X$ so $\frac {6\times 11}{13^3}$

Case II: The third ball is one of $W_1,W_2$. Then there are $6$ cases (two choices for the duplicate and three ways to place the odd man out). Thus $\frac 6{13^3}$.

Thus $$P(1,1)=\frac {6\times 11+6}{13^3}=\frac {72}{2197}$$

Consistency check: these should add to $1$. Indeed $$1331+397+397+72=2197$$