Vandermone's SRSWR | MStat 2017 PSB Problem 3

This is a beautiful problem form ISI MStat 2017 PSB Problem 3, where we use the basics of Bijection principle and Vandermone's identity to solve this problem.

Problem

Consider an urn containing 5 red, 5 black, and 10 white balls. If balls are drawn without replacement from the urn, calculate the probability that in the first 7 draws, at least one ball of each color is drawn.

Prerequisites

Algebrify the problem
Vandermone's Identity
${{n_{1}+\cdots+n_{p}} \choose m} = \sum_{k_{1}+\cdots+k_{p}=m} {n_{1} \choose k_{1}} \times {n_{2} \choose k_{2}} \times \cdots \times {n_{p} \choose k_{p}}$

Bijection Principle

Solution

It may give you the intuition, there is atleast in the problem, so let's do complementing counting. Okay! I suggest you to travel that path to help you realize the complexity of that approach.

Nevertheless, let's algebrify the problem.

You want to select $x_1$ red balls, $x_2$ blue balls, and $x_3$ white balls so that $x_1 + x_2 + x_3 = 7$ .

Now, $1 \leq x_1 \leq 5$ , $1 \leq x_2 \leq 5$ and $1 \leq x_3 \leq 10$ represents our desired scenario.

$0 \leq x_1 \leq 5$ , $0 \leq x_2 \leq 5$ and $0 \leq x_3 \leq 10$ denotes total number of cases.

Now, for each such triplet ( $x_1, x_2, x_3$ ) of the number of balls of each colour we have selected, we can select them in ${ 5 \choose x_1} \times {5 \choose x_2 } \times {10 \choose x_3}$ .

Let $P = \{ (x_1, x_2, x_3) | x_1 + x_2 + x_3 = 7, 0 \leq x_1 \leq 5, 0 \leq x_2 \leq 5, 0 \leq x_3 \leq 10 \}$ .

Total Number of Ways we can select the 7 balls =

$\sum_{(x_1, x_2, x_3) \in P} { 5 \choose x_1} \times {5 \choose x_2 } \times {10 \choose x_3} = { {5 + 5 + 10} \choose { x_1 + x_2+ x_3} } = { {5 + 5 + 10} \choose {7} }$

$Q = \{ (x_1, x_2, x_3) | x_1 + x_2 + x_3 = 7, 1 \leq x_1 \leq 5, 1 \leq x_2 \leq 5, 1 \leq x_3 \leq 10 \}$ .

Total Number of Ways we can select the 7 balls such that at least one ball of each color is drawn =

$\sum_{(x_1, x_2, x_3) \in Q} { 5 \choose x_1} \times {5 \choose x_2 } \times {10 \choose x_3}$

Let, $R = \{ (z_1, z_2, z_3) | z_1 + z_2 + z_3 = 4, 0 \leq z_1 \leq 4, 0 \leq z_2 \leq 4, 0 \leq z_3 \leq 9 \}$ .

Observe that $Q$ has a bijection with $R$ . $x_i = z_i +1$ .

Total Number of Ways we can select the 7 balls such that at least one ball of each color is drawn =

$\sum_{(x_1, x_2, x_3) \in R} { 4 \choose z_1} \times {4 \choose z_2 } \times {9 \choose z_3} = { {4 + 4 + 9} \choose { z_1 + z_2+ z_3} } = { {4 + 4 + 9} \choose {4} }$

Exercises

Prove the Vandermone's Identity.
Find the Probability.
Generalize the problem to general numbers and prove it.
Generalize the problem where the lower bounds of $x_i$ are $m_i$ .
What if there are upper bounds on $(x_1, x_2, x_3)$ ? [ Hint: Generating Functions ].

Stay Tuned! Stay Blessed!

This is a beautiful problem form ISI MStat 2017 PSB Problem 3, where we use the basics of Bijection principle and Vandermone's identity to solve this problem.

Problem

Prerequisites

Algebrify the problem
Vandermone's Identity
${{n_{1}+\cdots+n_{p}} \choose m} = \sum_{k_{1}+\cdots+k_{p}=m} {n_{1} \choose k_{1}} \times {n_{2} \choose k_{2}} \times \cdots \times {n_{p} \choose k_{p}}$

Bijection Principle