Conditional probability problem about dependent events?

Question

Sorry about the not-so-descriptive title. Here's the problem:

"There are $n$ coins arranged on a table, and $m$ of them are heads. You randomly select $p$ coins to remove one by one, but you forget whether the coins you removed were heads or tails. You then randomly pick a coin off of the table.

1. What is the probability that you choose tails?
2. Given that this coin is tails, what is the probability that at least one tail was removed among the first $p$?"

I've done some basic work but am largely stuck on this one. Here's what I have so far: The denominator for the final probability is going to be $n-p$. The probability of getting a tail, before accounting for coins falling off the table, is $\frac{n-m}{n}$.

My confusion in both parts is that the probability is changing at every step, isn't it? Each time one of the $p$ coins is removed, the probability of getting a tail isn't even just $\frac{n-m}{n-p}$, because the answer is entirely dependent on whether or not past coins removed were heads or tails.

I've not encountered a problem like this before. It seems to be a probability nested within a probability, and any help at all would be much appreciated.

Answer 1

"There are $n$ coins arranged on a table, and $m$ of them are heads. You select randomly select $p$ coins to remove one by one, but you forget whether the coins you removed were heads or tails. You then randomly pick a coin off of the table.

Without loss of generality, lets say you do this by first randomly arranging the coins in a line then move the first $p$ coins off the end of the line to the side.

What is the probability that you choose tails?

What is the probability that the $p+1$ coin in the line is a tail? Indeed, pick any place in line at random, then what is the probability that that coin is a tail?

Given that this coin is tails, what is the probability that at least one tails fell off the table?"

What is one minus the conditional probability that none of the first $p$ coins were tails when given that that coin was a tails.

Fix a tail in one position, and there are $n-1\choose m$ ways of arranging the $n-1$ other coins.

Likewise, fix $p$ heads and one tail and there are there are how many ways of arranging the how many remaining coins?

The ratio of these is the complement, so what you want is

$$1-\left({n-p-1 \choose m-p}\middle/{n-1\choose m}\right)$$

Answer 2

Hint : let's say n=7 and m=4 and p=2 Then, there are several possibilities :

-either you removed 1 head and 1 tail. In that case you have a probability 2/5 to choose tail. And the probability of having removed 1head and 1tail is $\frac{{{3}\choose{1}} {{4}\choose{1}}}{{7}\choose{2}}$

-either you removed 2heads. In that case you have a probability of 3/5 to choose tail. And the probability of having 2heads removed is $\frac{{{4}\choose{2}} }{{7}\choose{2}}$

-either you removed 2tails. In that case you have a probability of 1/5 to choose tail. And the probability of having 2tails removed is $\frac{{{3}\choose{2}} }{{7}\choose{2}}$

finally the probability is $\frac{2}{5}.\frac{{{3}\choose{1}} {{4}\choose{1}}}{{7}\choose{2}} + \frac{3}{5}.\frac{{{4}\choose{2}} }{{7}\choose{2}} + \frac{1}{5}.\frac{{{3}\choose{2}} }{{7}\choose{2}}$.

Try to understand it and generalize it.