Question

In Problem $4.15$, let team number $1$be the team with the worst record, let team number $2$be the team with the second-worst record, and so on. Let $Y_i$denote the team that gets the draft pick number $i$. (Thus, $Y_1=3$if the first ball chosen belongs to team number $3$.) Find the probability mass function of

(a) $Y_1$

(b) $Y_2$

(c)$Y_3$.

Short answer

(a)$\frac{12-i}{66}$

(b)$\sum _{j\ne i}\frac{12-i}{54+j}·\frac{12-j}{66}$

(c)$\sum _{k\ne j}\left[\sum _{j\ne i}\frac{11}{42+k+j}·\frac{12-k}{54+j}\right]·\frac{12-i}{66}$

Step-by-step solution

Given information Part (a)  

It is given that $66$balls placed in an urn such that 11 have the name of the team with the worst record, $10$have the name of the team with the $2^{\text{nd}}$worst record, and so on; finally, we have $1$ball with the ${11}^{\text{th}}$worst record.

That means, we have the balls in the following pattern.

$\begin{array}{|cc|}\hline \text{Team}& \text{Number of balls}\\ 1& 11(=12-1)\\ 2& 10(=12-2)\\ ⋮& ⋮\\ i& 12-i\\ ⋮& ⋮\\ 11& 1(=12-11)\\ \hline\end{array}$

Thus, team $i$consists of $12-i$balls.

Explanation Part (a)  

Find the probability mass function of $Y_1$.

Let $Y_i$denote the team that gets the draft pick number $i$. That means $Y_i$denotes the event of choosing team $i$.

From the given, we have $12-i$balls for the team $i$.

If we choose a team $1$, then $i=1$

Therefore, the probability mass function of $Y_1$is $P\left(Y_1=i\right)=\frac{12-i}{66}$

Given information Part (b)  

It is given that 66 balls are placed in an urn such that 11 have the name of the team with the worst record, 10 have the name of the team with the $2^{\text{nd}}$worst record, and so on; finally, we have $1$a ball with the ${11}^{\text{th}}$worst record.

That means, we have the balls in the following pattern.

$\begin{array}{|cc|}\hline \text{Team}& \text{Number of balls}\\ 1& 11(=12-1)\\ 2& 10(=12-2)\\ ⋮& ⋮\\ i& 12-i\\ ⋮& ⋮\\ 11& 1(=12-11)\\ \hline\end{array}$

Thus, the team $i$ consists of$12-i$balls.

Explanation Part (b)  

Find the probability mass function of $Y_2$.

In the $2^{\text{nd}}$draft pick $\left(Y_2=i\right)$, we have to select a team other than the team selected in the first draft pick.

That means, if we pick a team $i$in the $1^{\text{st}}$draft pick, then the $2^{\text{nd}}$draft pick is made from the remaining balls $66-(12-j)$(as we have $(12-i)$balls in the name of team $i$).

Thus, the probability of second draft pick is given as

$P\left(Y_2=i\right)=\sum _{j\ne i}P\left(\left\{Y_2=2\right\}\cap \left\{Y_1=i\right\}\right)$

$=\sum _{j\ne i}P\left(\left\{Y_2=2\right\}\mid \text{team}i\text{chosen first}\right)·P\left(\text{team}i\text{chosen first}\right)$

$=\sum _{j\ne i}\frac{12-i}{66-(12-j)}·\frac{12-j}{66}$

$=\sum _{j\ne i}\frac{12-i}{54+j}·\frac{12-j}{66}$

Therefore, the probability mass function of$Y_2\text{is}P\left(Y_2=i\right)=\sum _{j\ne i}\frac{12-i}{54+j}·\frac{12-j}{66}$

Given information Part (c)  

It is given that $66$balls are placed in an urn such that $11$have the name of the team with the worst record, $10$have the name of the team with the $2^{\text{nd}}$worst record, and so on; finally, we have 1 a ball with the ${11}^{\text{th}}$worst record.

That means, we have the balls in the following pattern.

$\begin{array}{|cc|}\hline \text{Team}& \text{Number of balls}\\ 1& 11(=12-1)\\ 2& 10(=12-2)\\ ⋮& ⋮\\ i& 12-i\\ ⋮& ⋮\\ 11& 1(=12-11)\\ \hline\end{array}$

Thus, the team $i$consists of $12-i$balls.

Explanation Part (c)  

Find the probability mass function of $Y_3$.

In the $3^{\text{rd}}$draft pick $\left(Y_3=i\right)$, we have to select a team other than the teams selected in the first two draft picks.

That means, if we pick a team $i$in the $1^{\text{st}}$draft pick and a team $j$in the $2^{\text{nd}}$draft pick, then the $3^{\text{rd}}$draft pick is made from the remaining balls $66-(12-k)-(12-j)$(as we have $(12-i)$balls in the name of team $i$and $(12-j)$balls in the name of team $j$).

Thus, the probability of third draft pick is given as

$P\left(Y_3=i\right)=\sum _{k\ne j}\sum _{j\ne i}P\left(Y_3=i,Y_1=j,Y_2=k\right)$

$P\left(Y_3=i\right)=\sum _{k\ne j}\sum _{j\ne i}P\left(Y_3=i,Y_1=j,Y_2=k\right)$

$=\sum _{k\ne j}\left[\begin{array}{l}\sum _{j\ne i}\left[\begin{array}{l}\left.P\left(\{X=3\}\mid \text{team}j\text{chosen}{2}^{\text{nd}}\&\text{team}i\text{chosen}{1}^{\text{st}}\right)\right]\\ \times P\left(\text{team}j\text{chosen}{2}^{nd}\mid \text{team}i\text{chosen}{1}^{\text{st}}\right)\end{array}\right]\\ \times P\left(\text{team}i\text{chosen first}\right)\end{array}\right]$

$=\sum _{k\ne j}\left[\sum _{j\ne i}\frac{11}{66-(12-k)-(12-j)}·\frac{12-k}{66-(12-j)}\right]·\frac{12-j}{66}$

$=\sum _{k\ne j}\left[\sum _{j\ne i}\frac{11}{42+k+j}·\frac{12-k}{54+j}\right]·\frac{12-i}{66}$

Therefore, the probability mass function of $Y_3$is

$P\left(Y_3=i\right)=\sum _{k\ne j}\left[\sum _{j\ne i}\frac{11}{42+k+j}·\frac{12-k}{54+j}\right]·\frac{12-i}{66}$