A question on fundamentals of Probability and Random Variables Suppose you own 6 shirts. Starting April 1st, you pick a shirt uniformly at random to wear each day. The choice of shirt is independent of one another. Assume that a shirt worn on a given day is laundered at night, and is available to be chosen the next day.
(a) Given that you happened to select your third distinct shirt on April 10th, how many further days would you have to wait on an average, until you select your fourth distinct shirt?
(b) Find the expected total number of days that would elapse(starting April 1st), until you have worn each of the six shirts at least once?
For the part (a), I modelled the day when he wore the 2nd distinct shirt as Binomial Distribution, but unable to proceed forward. Kindly help me in solving this question. Thanks in advance.
 A: (a) Suppose that on average you have to wait $\mu$ days. Then we have the equality:$$\mu=\frac36\cdot1+\frac36\cdot(1+\mu)\tag1$$
The first $\frac36$ stands for the probability that on April the $11$-th you pick out a shirt distinct from the $3$ that were used already. In that case you have to wait $1$ day exactly. That's why the second factor of the first term is $1$.
The second $\frac36$ stands for the probability that on April the $11$-th you pick out a shirt not distinct from the $3$ that were used already. In that case you arrive in the same position as you were at April the $10$-th except that you are $1$ day further. That's why the second factor of the second term is $1+\mu$.
Based on $(1)$ we find easily that $\mu=2$.
(b) As in (a) we can find the averages/expectations also in the more general case that you have worn $k\in\{0,1,\dots,5\}$ distinct shirts already (instead of $3$ as in (a)). That gives similarly the equality:$$\mu_k=\frac{6-k}{6}\cdot1+\frac{k}{6}\cdot(1+\mu_k)=1+\frac{k}6\cdot\mu_k$$ leading to: $$\mu_k=\frac6{6-k}$$
Hence the total expected waiting time is:$$\sum_{k=0}^5\frac6{6-k}=6\sum_{k=1}^6\frac1k$$
A: For a) we use the geometric distribution rather than the binominal distribution. The probability to select one of the three (distinct) shirts which has already been  worn is $\frac36=\frac12$. Therefore the probability to select your fourth distinct shirt on the n-th day is $P(X=n)=\left(\frac{1}{2}\right)^{n-1}\cdot \frac{1}{2}$. Thus the expected value is ...
For b) you can read the nice answers to this similar question. I´m pretty sure that you can make the transfer to answer your question. 
