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An urn contains 4 white balls and 6 red balls. Balls are drawn from the urn one at a time without replacement. The random variable Xi is 1 if the i-th ball drawn is red, and 0 otherwise.

a) What is the joint probability mass function of X1 and X2?

b) What is the joint probability mass function of X1,X2 and X3?

So where I'm stuck is in identifying if X1 is when red or if X2 is red? I mean logically I would go from X1 to X10 and get the total probability but I don't think I have to go that far for this question. Just not sure how I should start.

If it is the way I think, then my calculation for a would be: $P(x_1 = 1, x_2 = 1) = 6/10 * 5/9$

$P(x_1 = 1, x_2 = 0) = 6/10 * 4/9$

$P(x_1 = 0, x_2 = 1) = 4/10 * 6/9$

$P(x_1 = 0, x_2 = 0) = 4/10 * 3/9$

And b) would follow a similar pattern. But this time X3 would be involved as well.

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I will give an explanation for part $(1)$ and similarly you can do the second part.
CASE $(1)$: We must enumerate every combination of $0/1$ values for $X_1$ and $X_2$ and determine the combination of draws that result in those values. First consider
$p(0, 0): X_1 = 0, X_2 = 0$. This means that the first ball drawn is not red and the second ball drawn is not red. This means both balls drawn are white. Initially, there are 4 white balls in the urn and 10 balls totally. On the second draw, there are 3 red balls and 9 balls total. So, $ p(0, 0) = \frac{4}{10}\times \frac{3}{9} =\frac{12}{90}$ Note that the denominator will always be the same since on each draw we draw one ball, so I will exclude this from my derivation.
$p(0,1) : X_1 = 0,X_2 = 1$ means that the first ball is not red and the second ball is red. Initially there are 4 white balls. On the second draw, we draw a red ball of which there are 6 since none of them had been drawn yet. So $p(0, 1) = \frac{4}{10}\times \frac{6}{9} = \frac{24}{90}$.
$p(1,1) : X_1 = 1, X_2 = 1$ means that the first ball is red and the second ball is red. Initially there are 6 red balls in the urn. On the second draw, after one of the red balls is chosen, there are 4 remaining. So $p(1,1) = \frac{6}{10}\times \frac{5}{9} = \frac{30}{90}$.
$p(1,0) : X_1 = 1,X_2 = 0$ means that the first ball is red and the second ball is white. Initially there are 6 red balls. On the second draw there are 4 white balls. So $p(1,0)= \frac{6}{10}\times \frac{4}{9} = \frac{24}{90}$.

Similarly for CASE $(2)$, consider the probabilities $p(0,0,0) ; p(0,0,1) ; p(0,1,0) ; p(1,0,0) ; p(1,1,0) ; p(1,0,1) ; p(0,1,1) ; p(1,1,1) $ and proceed similarly.

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  • $\begingroup$ Ahh thanks, this is exactly what I was thinking. Makes perfect sense too. I managed to figure out part b as well. $\endgroup$ – quwerty Dec 7 '16 at 10:54
  • $\begingroup$ Also for part b I would also consider p(0,1,0) as well correct? You seem to have missed that one. $\endgroup$ – quwerty Dec 7 '16 at 10:55
  • $\begingroup$ No, that is also included. See the third term. $\endgroup$ – Rohan Dec 7 '16 at 10:55
  • $\begingroup$ Ah sorry, didn't notice that. Thanks! $\endgroup$ – quwerty Dec 7 '16 at 10:58
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Yes, $p_{_{X_1,X_2}}(1,1) = \frac{6}{10}\frac{5}{9} \\p_{_{X_1,X_2}}(1,0) = \frac{6}{10}\frac{4}{9} \\p_{_{X_1,X_2}}(0,1) = \frac{4}{10}\frac{6}{9} \\p_{_{X_1,X_2}}(0,0) = \frac{4}{10}\frac{3}{9}$

Which can be summarised using $\newcommand{\permut}[2]{{^{#1}{\mathrm P}_{#2}}} \permut n k =n^{\underline k} =n!/(n-k)! = n(n-1)\cdots(n-k+1)$

$${p}_{_{X_1,X_2}}(x,y) = \dfrac{\permut 4{2-x-y}~\permut 6{x+y}}{90}\mathbf 1_{x\in\{0,1\},y\in\{0,1\}}$$

And likewise to save typing out the eight cases for the trivariate joint probability.

$${p}_{_{X_1,X_2,X_3}}(x,y,z) = \dfrac{\permut 4{3-x-y-z}~\permut 6{x+y+z}}{720}\mathbf 1_{(x,y,z)\in\{0,1\}^3}$$

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