The answer to the first question is correct.

What is the initial probability p(nS) that the bag contains nS silver coins?

$$P(nS) = 1/11 $$

Now, the second question

You pull out one coin: it is silver. What is now the probability that the bag contains 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 silver coins?

The analysis in the question is wrong because the question is about conditional probability. We want to find out the probability of the event that the bag contains n silver coins($0 \le n \le 9$), **given** that the first coin pulled out was a silver coin. We will use Bayes' theorem to solve this.

See this and this for Bayes theorem and this for an illustrative example. The idea of drawing all the probabilities as a tree demonstrated in this idea is quite helpful in understanding the problem. Now, solution to the problem.

Lets us call the event we pulled out one silver coin as $O$.

Let us call the event there are n silver coins remaining in the bag as $nS$, where $0 \le n \le 9$

We want to find out $P(nS | O)$

Using Bayes formula

$$P(nS|O) = \frac{P(O|nS). P(nS)}{P(O)}$$

From the above question, we know that

$$P(nS) = 1/11 $$

$P(O|nS)$ is defined as the probability where a silver coin is picked given there are n silver coins remaining in the bag after the ball is picked. Since there are n silver coins remaining after the first silver coin was picked, the probability $P(O|nS)$ can be defined as $\frac{n+1}{10}$

$$P(O|nS) = \frac{n+1}{10}$$

$P(O)$ is the probability that a silver coin is picked from the bag on the first turn. Since there are either silver or gold coins in the bag, there is equal probability for both. Thus,

$$P(O) = 1/2$$

This can also be derived by drawing the tree as illustrated in the video

$$P(O) = (\frac{1}{11}*\frac{0}{10}) + (\frac{1}{11}*\frac{1}{10}) + .... + (\frac{1}{11}*\frac{9}{10}) + (\frac{1}{11}*\frac{10}{10})$$

$$P(O) = \frac{1}{11}(\frac{0}{10} + \frac{1}{10} + .... + \frac{9}{10} + \frac{10}{10})$$

$$P(O) = \frac{1}{11}(\frac{0+1+2+ ... + 9 + 10}{10}) = \frac{55}{11*10} = \frac{1}{2}$$

Substituting these values in

$$P(nS|O) = \frac{P(O|nS). P(nS)}{P(O)}$$

$$P(nS|O) = \frac{n+1}{55}$$

You now pull out another coin: it is gold. What is now the probability that the bag contains 0, 1, 2, 3, 4, 5, 6, 7, 8 silver coins?

Let us call the event where two coins were pulled, and the first one is silver and other is gold as $T$.

Let us call the event there are n silver coins remaining in the bag as $nS$, where $0 \le n \le 8$

We want to find out $P(nS | T)$

Using Bayes formula

$$P(nS|T) = \frac{P(T|nS). P(nS)}{P(T)}$$

We know that

$$P(nS) = 1/11 $$

$P(T|nS)$ is defined as the probability where a silver coin is drawn first and gold coin is drawn next given there are n silver coins remaining in the bag after the two coins are picked. On the first turn, the probability for a silver coin to be picked is $\frac{n+1}{10}$. In the second turn, 9 coins would be left in the bag and n of them are silver coins, so, there would be $9-n$ gold coins. The probability of picking a gold coin in the second turn would be $\frac{9-n}{9}$. Combined, the probability would be

$$P(T|nS) = \frac{n+1}{10}*\frac{9-n}{9}$$

$P(T)$ is the probability that a black ball is picked from the bag first followed by a white ball. This can be derived from the probability tree

$$P(T) = (\frac{1}{11}*\frac{0}{10}) + (\frac{1}{11}*\frac{1}{10}*\frac{9}{9}) + (\frac{1}{11}*\frac{2}{10}*\frac{8}{9}) .... + (\frac{1}{11}*\frac{9}{10}*\frac{1}{9}) + (\frac{1}{11}*\frac{10}{10}*\frac{0}{9})$$

$$P(T) = \frac{1}{11}(\sum_{n=0}^{10} \frac{n(10-n)}{10*9})$$

$$P(T) = \frac{1}{11}(\frac{165}{90}) = \frac{1}{6}$$

Substituting these values in

$$P(nS|T) = \frac{P(T|nS). P(nS)}{P(T)}$$

$$P(nS|T) = \frac{8n+9-n^2}{165}$$

## Best Answer

Your general reasoning/intuition is correct, but your calculations are not.

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

unconditionalprobability 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.