# bag of 20 marbles, at least 19 are white, 20th is red with prob. P. What is likelihood of finding red on successive draws?

I have bag of 20 marbles, at least 19 of which are white, 20th marble may be red with prob. P. I plan to draw marbles one at a time till I find the red marble or all marbles are used up.

At start, I believe I can say that probability that I will find a red marble in this bag is P. However, as I draw each white marble how does this probability change? (I would stop as soon as I find the red marble, but there may be no red marble in the bag.)

Ideally, I would like to know the probability that I will find the red marble on k'th draw where k = 1..20 given that I have not found it in any previous draws.

Let $A$ be the event that the bag contains a red marble, $B_k$ be the event that you picked white marbles in each of the first $k$ draws.

The first quantity that you want: $Pr[A|B_k]$. As you observed, this value is just $Pr[A]=p$ when $k=0$.

$Pr[A|B_k]$ is equal to $\dfrac{Pr[A \cap B_k]}{Pr[B_k]}=\dfrac{Pr[A]Pr[B_k|A]}{Pr[B_k]}=\dfrac{pPr[B_k|A]}{Pr[B_k]}$.

The two probabilities in the numerators and denominator are easy to calculate.

The second quantity, that is the probability of finding the red marble in the $k$th draw, is simpler: it is equal to $p$ times $q$, where $q$ is the probability that the first $k-1$ marbles are white and the $k$th one is red, assuming that the bag has one red marble.

The probability that you find the red marble on draw $k$ is $p$ (that there is a red marble) times $\frac 1{20}$ or $\frac p{20}$ The chance that you draw $k-1$ white marbles first is $(1-p)+p\frac {21-k}{20}$ because if there is a red marble you can line them up in the order you will draw them and get $k-1$ white ones as long as the red is in position $k$ or later. The chance you get a red on draw $k$ given that you got $k-1$ whites is then the ratio of these $$\frac {\frac p{20}}{(1-p)+p\frac {21-k}{20}}$$

Hint: What is the probability that, by the $k$th draw, you still have not encountered a red marble?

Fuller answer: If $A_k$ is the event that you do not draw a red ball in the first $k$ draws, and $B$ is the probability that there is no red ball, you want:

$$P(B\mid A_k)=\frac{P(B\cap A_k)}{P(A_k)}$$

But $P(B\cap A_k)=1-P$, since when there is no red ball, $A_k$ is always true.

And: $$P(A_k)=1-P + P\frac{\binom{19}{k}}{\binom{20}{k}}=1-P+P\frac{20-k}{20}=1-\frac{Pk}{20}$$

So the probability there is no red ball if you haven't found one in $k$ draws is:

$$\frac{1-P}{1-\frac{Pk}{20}}=\frac{20-20P}{20-kP}$$

The probability of there being a red if not found in $k$ draws is:

$$P(\lnot B\mid A_k)=1-P(B\mid A_k)=1-\frac{20-20P}{20-kP}=\frac{(20-k)P}{20-kP}$$