Draws without replacement

Idea: Order is your choice

Draws without replacement depend on each other, but you may count them in order or as a hand.

• Both routes agree if you stay consistent
• Use the same convention on the top and the bottom

Example: Exactly 2 red of 3

An urn has $5$ red and $4$ blue. Draw $3$ without replacement. Chance of exactly $2$ red?

1. Hands: $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$
2. Favourable: $\left(\genfrac{}{}{0ex}{}{5}{2}\right)\times \left(\genfrac{}{}{0ex}{}{4}{1}\right)=40$
3. Chance: $\frac{40}{84}=\frac{10}{21}$

Idea: The ordered route

One order, red red blue, has chance $\frac{5}{9}\times \frac{4}{8}\times \frac{4}{7}=\frac{10}{63}$.

• Every order has the same chance
• $3$ places for the blue ball: $3\times \frac{10}{63}=\frac{10}{21}$

Formula: Exactly k red

$P(k\text{red})=\left(\genfrac{}{}{0ex}{}{r}{k}\right)\times \frac{\left(\genfrac{}{}{0ex}{}{b}{d-k}\right)}{\left(\genfrac{}{}{0ex}{}{r+b}{d}\right)}$

r red, b blue, d draws without replacement. For several k, add the cases.

Common mistake: Mixing conventions

$\frac{10}{63}$ is the chance of one particular order, not of $2$ red in any order.

Instead Decide ordered or unordered once, and use it on the top and the bottom.

Idea: Every position is alike

Before anything is revealed, each position in a shuffle has the same chance.

• Second card an ace: $\frac{4}{52}=\frac{1}{13}$
• Second card a heart, so the first is a heart with $\frac{12}{51}=\frac{4}{17}$
• Mean red count: $3\times \frac{5}{9}=\frac{5}{3}$

Quick check: All the same colour

An urn has $5$ red and $4$ blue. Draw $3$ without replacement. Chance all $3$ are the same colour?

Idea: With replacement

Putting balls back makes the draws independent.

• $2$ red in $3$ draws: $3\times \frac{5}{9}\times \frac{5}{9}\times \frac{4}{9}=\frac{100}{243}$
• Against $\frac{10}{21}\approx 0.48$ without replacement
• The mean stays $\frac{5}{3}$, but without replacement keeps closer to it

Idea: Say it out loud

Say which convention you picked and why.

• Give the size of the list, then the favourable count
• If you have time, confirm with the other route
• Name symmetry when you use it