# Probability question involving 4 coins (gold and silver) in 3 boxes each. Is the explanation correct

conditional probabilityprobability

Key:
g = Gold coin.
s = Silver coin.

Box A)
g1
g2
g3
g4
Box B)
g5
g6
s5
s6
Box C)
s1
s2
s3
s4

Waz chooses a box at random and picks up a coin from it. It turns out to be a gold coin.
( Pick 1 from Box ? results in: g being picked )

Question 1:
Which box do you think g (pick 1) is more likely to have been picked from?

Box C is impossible. Box A has more g than Box B, so Box A is more likely.
Box A: 4/8 or 1/2
Box B: 2/8 or 1/4

Question 2: For pick 2 from that same Box ?, which do you think is more likely to be picked by Waz: g or s?

Pick 1 establishes Box C is impossible leaving 8 possibilities. However, since one g has been taken out of the equation, this now leaves only 7 possibilities.
5/7 = g and 2/7 = s.

Hence pick 2 is more likely to be g than s.

Is my probability calculation correct here? (bolded parts)

A1: Box A: 1/2 , Box B: 1/4
A2: I think it's either 5/7 = g or 4/6 = g and 2/7 = s or 2/6 = s.
I'm not sure.
Are there only 7 possibilities or 6 remaining for pick 2? Or something else?

Let $$A$$, $$B$$, $$C$$ represent the events that the box selected was $$A$$, $$B$$, or $$C$$ respectively. Let $$G$$ represent the event that the first drawn coin was gold.

Then we are asked to compute and compare $$\Pr[A \mid G]$$, $$\Pr[B \mid G]$$, and $$\Pr[C \mid G]$$; these are the conditional probabilities that the selected box was $$A$$, $$B$$, or $$C$$ given the drawn coin was gold. We have from Bayes' rule

$$\Pr[A \mid G] = \frac{\Pr[G \mid A]\Pr[A]}{\Pr[G]}.$$

Since before drawing a coin from the box, each box is equally likely to have been selected, we know $$\Pr[A] = \Pr[B] = \Pr[C] = 1/3$$. We also know that $$\Pr[G \mid A] = 1$$ because every coin in box $$A$$ is gold. The only unresolved quantity is $$\Pr[G]$$, the unconditional probability of drawing a gold coin. Our intuition suggests that since among all three boxes, there are an equal number of gold and silver coins, that we should have $$\Pr[G] = 1/2$$. Indeed, this is true as the law of total probability demonstrates:

\begin{align} \Pr[G] &= \Pr[G \mid A]\Pr[A] + \Pr[G \mid B]\Pr[B] + \Pr[G \mid C]\Pr[C] \\ &= (1)(1/3) + (1/2)(1/3) + (0)(1/3) \\ &= 1/2. \end{align}

So the desired probability is $$\Pr[A \mid G] = \frac{1/3}{1/2} = \frac{2}{3}.$$ Similarly,

$$\Pr[B \mid G] = \frac{\Pr[G \mid B]\Pr[B]}{\Pr[G]} = \frac{(1/2)(1/3)}{1/2} = \frac{1}{3},$$

and of course

$$\Pr[C \mid G] = \frac{\Pr[G \mid C]\Pr[C]}{\Pr[G]} = \frac{(0)(1/3)}{1/2} = 0.$$

This formally proves that the most likely box from which the gold coin was drawn is box $$A$$, but the probability is $$2/3$$, not $$1/2$$ as you asserted.

For the second coin drawn from the same box, again intuition suggests that this coin should also be gold, since $$A$$ was the more likely box (twice as likely as $$B$$) that was selected. If we let $$G'$$ be the event that the second coin drawn (without replacement) is gold, then we are asked to compute $$\Pr[G' \mid G]$$, the probability of the second coin being gold given that the first coin was gold. Thus,

$$\Pr[G' \mid G] = \Pr[G' \mid A, G]\Pr[A \mid G] + \Pr[G' \mid B, G]\Pr[B \mid G].$$ We omit box $$C$$ since $$\Pr[C \mid G] = 0$$. Then

$$\Pr[G' \mid A, G] = 1,$$

since if the first gold coin was drawn from box $$A$$, the second coin will necessarily be gold. But

$$\Pr[G' \mid B, G] = \frac{1}{3},$$

since if the first gold coin was drawn from box $$B$$, there remains only $$1$$ gold coin out of $$3$$ in that box. Therefore,

$$\Pr[G' \mid G] = (1)(2/3) + (1/3)(1/3) = \frac{7}{9}.$$

Note how the calculation must match the sampling process. Your calculation is incorrect because you seem to be allowing the second coin to be selected from a different box as the first, when the problem clearly states that once the first coin is drawn, the second draw is from the same box.