We roll a die until we get $4$ fives. Let $X$ be the number of rolls needed for the first $5$ and let $Y$ be the number of rolls needed to get the fourth five. What is the joint probability mass function of $X,Y.$

My attempt: $P(X=x \cap Y=y)=$ ${y-1}\choose{3}$$\left({\frac{1}{6}}\right)^4$$\left({\frac{5}{6}}\right)^{y-4}$

My reasoning: There are ${y-1}\choose{3}$ ways of choosing 3 spots for fives among the first $y-1$ rolls. Then we have $\left({\frac{1}{6}}\right)^4$= probability of rolling four fives. And $\left({\frac{5}{6}}\right)^{y-4}$= probability of rolling $y-4$ non-fives.

Does this look correct?

  • $\begingroup$ You can find unique marginal probabilities if you're given a joint distribution, but without some extra information, the joint distribution cannot be uniquely determined. See, e.g., copulas en.wikipedia.org/wiki/Copula_(statistics) $\endgroup$ – user99680 Dec 6 '13 at 5:08
  • $\begingroup$ Oh, I see, I guess my comment applies only for continuous distributions, I guess. $\endgroup$ – user99680 Dec 6 '13 at 7:42

You need to do it in this way . Let $x$ be the number of tries required to get the first $5$ then $p_X(x) = (\frac{5}{6})^{x-1}\frac{1}{6}$ After this you need $3$ more $5$ . You can treat this as a completely new event , because it doesn't depend on what happened earlier .

Now let $y$ be the total number of trials needed to get four $5$ , then you have already wasted $x$ tries . You are left with only $y-x$ tries now . Now the ways of getting two $5$s will be ${y-x-1}\choose{2}$ . Therefore the probability of this event will be ${y-x-1}\choose{2}$$(\frac{1}{6})^2 (\frac{5}{6})^{y-x-3}(\frac{1}{6})$

Therefore the joint PMF $p_{X,Y}(x,y)$ = $(\frac{5}{6})^{x-1}\frac{1}{6}$${y-x-1}\choose{2}$$(\frac{1}{6})^2 (\frac{5}{6})^{y-x-3}(\frac{1}{6}) $


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