# Probability-envelope and coffers

Two chests contain very similar envelopes and each envelope contains either a $\$ 5$, a$\$10$ or a $\$ 20$. In the first chest, there are$6$envelopes each containing$\$5$ and another 6 containing $\$ 10$. In the second chest, there are$2$envelopes with each$\$5$ and $4$ envelopes of $\$ 20$. a)What is the probability that if a chest and an envelope are chosen at random, an envelope will contain a$\$5$?

b)What is the probability that the envelope comes from the first chest since it contains a $\$ 5$? For part$a$, I thought of maybe doing$(\frac 12 \times \frac 6{12}) + (\frac 12 \times \frac 26)$but i'm really not sure. For part$b$, I have no clue. Can someone please help me? Thank you !! • Your proposal for part a looks good. For$b$, use Bayes' theorem. – lulu Mar 9 '17 at 19:23 • Or simpler, having done (a), just the definition of conditional probability in (b). In abbreviated notation:$P(I|5) = P(I\cap 5)/P(5) = ??\$ You already have both the numerator and denominator from part (a). – BruceET Mar 9 '17 at 21:57