Continually collecting coupons - geometric distribution Suppose you conntinually collect coupons and that there are two different types of coupon, type A and type B. Suppose also that each time a new coupon is obtained it is a type A coupon with probability 1/3 and a type B coupon with probability 2/3, independently of what coupons you have collected so far. Let X be the number of coupons collected until you have at least one coupon of both types. a)find the pmf of X b) find E[X] I'm stumped on how to solve this question as I've seen the ones like it with each coupon type having an equal chance of being chosen and an unknown amount of types. I think its a geometric distribution, but I cannot figure out the pmf.

_Condition_ on the first coupon you get. _Given_ it is Type A, then $X-1$ has geometric distribution with parameter $2/3$. Thus, given the first coupon is of Type A, we have $X=j$ with probability $(1/3)^{j-2}(2/3)$, $j=2,3,4,\dots$.

Similarly, given the first coupon is of Type B, $X=j$ with probability $(2/3)^{j-2}(1/3)$. Thus $$\Pr(X=j)=(1/3)(1/3)^{j-2}(2/3)+(2/3)(2/3)^{j-2}(1/3)$$ for $j=2,3,4,\dots$.

For the expectation, we could write down the expression for the expectation as an infinite sum, and evaluate. But it is nicer to do a conditional expectation calculation, for then we can use the known expectation of a random variable with geometric distribution.

Given the first coupon is of Type A, the expected _additional_ time until we get a coupon of Type B is $\frac{3}{2}$, so the expected total time is $1+\frac{3}{2}$. Similarly, given the first is of Type B, the expected total time is $1+3$. Thus $$E(X)=\frac{1}{3}\left(1+\frac{3}{2}\right)+\frac{2}{3}\left(1+3\right).$$