What is the probability that the first ball drawn was black, given the event $X_3 \leq 5$

Question

Im trying to solve the following question:

There is a urn with 1 black ball and 1 red ball. A ball is drawn at random from the urn, it is placed back in the urn along with 2 more balls with the same color as the ball that was drawn. Denote the random variable Xi to be the number of black balls after the i-th draw, i = 1, 2, . . .. Note that $X_i$ is the random variable for number of black balls in the urn including the two new balls added after the i-th draw. What is the probability that the first ball drawn was black, given the event $X_3 \leq 5$?

This is what I have so far:

$$P(X_3=1 | X_3\leq 5)=\frac{P(X_3=1 \cap X_3\leq 5)}{P(X_3\leq 5)}=\frac{P(X_3=1)}{P(X_3=1)+P(X_3=3)+P(X_3=5)}$$

Hence,

$$\frac{\frac{5}{16}}{\frac{5}{16}+\frac{3}{16}+\frac{3}{16}}$$

Correct?

Answer 1

With a decision tree, the process is long, it is certainly not optimal, but it is always a good tool to obtain the result.

We collect all options (8 options)

1st level : P(B)=1/2 , P(W)=1/2

2nd level : $P(BB)=(1/2)\times(3/4)$, $P(BW)=(1/2)\times(1/4)$,$P(WB)=(1/2)\times(1/4)$, $P(WW)=(1/2)\times(3/4)$

3rd level : $P(BBB)=(1/2)\times(3/4)\times(5/6)$, $P(BBW)=(1/2)\times(3/4)\times(1/6)$

$P(BWB)=(1/2)\times(1/4)\times(3/6)$, $P(BWW)=(1/2)\times(1/4)\times(3/6)$

$P(WBB)=(1/2)\times(1/4)\times(3/6)$, $P(WBW)=(1/2)\times(1/4)\times(3/6)$

$P(WWB)=(1/2)\times(3/4)\times(1/6)$, $P(WWW)=(1/2)\times(3/4)\times(5/6)$

We know that the result is not BBB. So, we exclude this row.

Total of remaining values is $1-P(BBB)=11/16$

Total of 'success-values' is $P(BBW)+P(BWB)+P(BWW)= 3/16$

Probability is $(3/16)/(11/16) = 3/11$

Answer 2

$X_3$ can only be $>5$ (indeed $7$) if black is drawn all the way, which happens with probability $$\frac12×\frac34×\frac56=\frac5{16}$$ There is a $\frac12$ chance of the first draw being black and from there $X_3>5$ happens with probability equal to the last two terms above, i.e. $\frac58$. Then the final answer is $$\frac{(1/2)(1-5/8)}{1-5/16}=\frac3{11}$$

Answer 3

You calculated $P(X_3=1\mid X_3\le5)$ correctly, but that’s not what the questions asks for – it asks for $P(X_1=3\mid X_3\le5)$.

That the first ball drawn is black and $X_3\le5$ happens if the first ball drawn is black, with probablity $\frac12$, and then the second ball drawn is red, with probability $\frac14$, or it’s black, with probability $\frac34$, but then the third ball drawn is red, with probability $\frac16$.

That $X_3\le5$ happens unless $X_3=7$, with probability $\frac12\cdot\frac34\cdot\frac56$. Thus

\begin{eqnarray} P(X_1=3\mid X_3\le5) &=& \frac{P(X_1=3\cap X_3\le5)}{P(X_3\le5)} \\ &=& \frac{\frac12\left(\frac14+\frac34\cdot\frac16\right)}{1-\frac12\cdot\frac34\cdot\frac56} \\ &=& \frac{\frac3{16}}{\frac{11}{16}} \\ &=& \frac3{11}\;. \end{eqnarray}

Because multiple wrong answers (including my own initial answer) were given, I wrote this Java code to double-check the result by simulation. It confirms the result $\frac3{11}$.