probability - two gifts behind five envelopes (game)

Question

This question is from MIC-OCW 6.041 (Spring 2006), Q1 - tutorial-1, with this solution.

... a student is told that there is a single gift certificate for a sundae in one of 5 envelopes, say envelope 1, 2, 3, 4 and 5. The student picks an envelope, say $i$, but does not open it. The professor then opens another envelope, say $j$ where $j \ne i$, which the professor knows for sure does not contain the gift certificate.

1. Assuming that a certificate was equally likely to be in any envelope initially, what is the probability, after the professor showed an empty envelope, that a certificate is in envelope $k$, where $k = i$ and $k \ne j$ (i.e. the certificate is in the envelope that the student originally picked)?
2. Repeat part (1), for $k \ne i$ and $k \ne j$ (i.e. the certificate is in envelope k which the student did not pick and the professor did not open.)
3. Repeat (1) and (2), now assuming that there are gift certificates in two out of the five envelopes.

I am trying to solve part-3 of the above question, i.e. when two of the five envelopes have gifts in them. I used the following reasoning:

Notation: let $i$ be the envelop that the student picks. And, note that the professor never opens envelop-$i$.

• $I$ - event that the envelop $i$ contains the gift
• $\neg I$ - event that the envelop $i$ does not contain the gift
• $K$ - event that the envelop $k$ contains the gift, for some envelop $k \ne i$
• $\neg K$ - event that the envelop $k$ does not contain the gift, for some envelop $k \ne i$
• $P_k$ - event that the professor opens envelop $k$
• $\neg P_k$ - event that the professor opens an envelop other than $k$

See this .

Part-3.1: I did the following,

$$ \begin{align} P(I|P_j) & = \frac{P(I \cap P_j)}{P(P_j)} \\ & = \frac{P(I \cap P_j)}{P(I\cap\neg J \cap P_j) + P(\neg I \cap \neg J \cap P_j)} \\ & = \frac{\frac{2}{5}\frac{3}{4}\frac{1}{3}}{\frac{2}{5}\frac{3}{4}\frac{1}{3} + \frac{3}{5}\frac{1}{2}\frac{1}{2}} \\ & = \frac{2}{5} \qquad\text{// this is wrong!} \end{align} $$

I think there must be something wrong with how I'm computing the conditional-probabilities. So, what is wrong with my computations? And, how to solve this problem?

Answer 1

Way to solve 3):

Let $I$ denote the event that a prize is in envelop $i$.

Let $K$ denote the event that a prize is in envelop $k$.

If $k=i$ then: $$P(K)=P(I)=\frac25$$

If $k\notin\{i,j\}$ then: $$P(K)=P(K\mid I)P(I)+P(K\mid I^c)P(I^c)=\frac13\frac25+\frac23\frac35=\frac8{15}$$

If there is a gift in envelop $\{i\}$ then there is a gift in exactly $1$ of the $3$ other envelopes not opened by the professor. If there is no gift in envelop $\{i\}$ then there is a gift in exactly $2$ of the $3$ other envelopes not opened by the professor.