Probability of the last remaining item after removing items Question:
Mr. J has 3 shirts and he wears them each day with the following probabilities: Green (1/3), White(1/6) and Red(1/2).
Once he decided to clean out his closet, so each evening he independently decides with a prob. 1/5 to throw the shirt he was wearing that day. Each morning he picks out a shirt of the remaining shirts, according to the the same probabilities. For example if he threw the green shirt, then he will wear the red shirt with prob 3/4 and the white with prob 1/4.
What is the probability that the last shirt remaining in his closet is the white one?
a. 0.33
b. 0.58
c. 0.69
d. 0.83
Our thoughts
Me and my buddy are pretty frustrated from this one... We saw some solution that ignores the throwing out part completely. We were wondering why is that correct? And could anyone post a fully explained solution to this. Thanks in advance.
 A: Yes, you can indeed ignore the throwing out part as this is not at all dependent on which shirt he's wearing. You can think of this probability as essentially prolonging the process since we don't always throw out a shirt, but if we don't throw out a shirt then we start the next day with exactly the same probabilities as for the day before- nothing has changed. It is as if the day that we didn't throw a shirt out on didn't even happen. Hence we only really care about the days when we do throw out shirts and for those days we only really care about which shirt he was wearing that day based on the probabilities of picking each shirt. I think this should be enough for you to solve the problem now.
A: john is correct.  As he says, you're only interested in the days when shirts are thrown out, so the probability only affects how long it takes Mr. J to have only one shirt left.  So the probabilities for which shirts are left are:
$$
\begin{align}
p_{white \ left} = \frac{1}{3}\frac{3}{4}+\frac{1}{2}\frac{2}{3}=\frac{7}{12} \\
p_{red \ left} = \frac{1}{3}\frac{1}{4}+\frac{1}{6}\frac{2}{5}=\frac{3}{20} \\
p_{green \ left} = \frac{1}{6}\frac{3}{5}+\frac{1}{2}\frac{1}{3}=\frac{4}{15}
\end{align}
$$
These probabilities add up to 1.  The white is most likely to be left because it's worn with the lowest probability, so it's least likely to be thrown out.  $\frac{7}{12}$ is 0.58, so it's answer (b).
