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 !!

  • 1
    $\begingroup$ Your proposal for part a looks good. For $b$ , use Bayes' theorem. $\endgroup$ – lulu Mar 9 '17 at 19:23
  • $\begingroup$ 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). $\endgroup$ – BruceET Mar 9 '17 at 21:57

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