# Probability Ratio using Combination

### Counting Techniques for Probability Ratio using Combination

The probability ratio of an event is the likelihood of chance that the event will occur due to a random experiment, and it can be found using the combination. When the number of possible outcomes of a random experiment is infinite, the enumeration or counting of the sample space becomes tedious. In these situations, we make use of what has known as the counting technique principle for finding the probability. Combinations will be used if the order of the events is not significant.
$\require{color}$
$$\binom{n}{k} = \dfrac{n!}{(n-k)!k!}$$
Let’s take a look at the worked examples now.

### Probability using Combination

A bag contains $n$ pink cards, $n$ yellow cards and $n$ green cards. Three cards are drawn randomly from the bag, one at a time, without replacement.

(a)   Find the probability, $P_s$, that the three cards are all the same colour.

It could be three pink cards, three yellow cards, or three green cards out of $3n$ cards.
$\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}$
The probability that the three cards are all pink is $\displaystyle \ddfrac{\binom{n}{3}}{\binom{3n}{3}}$.
The probability that the three cards are all yellow is $\displaystyle\ddfrac{\binom{n}{3}}{\binom{3n}{3}}$.
The probability that the three cards are all green is $\displaystyle\ddfrac{\binom{n}{3}}{\binom{3n}{3}}$.
$\therefore P_s = \displaystyle\ddfrac{\binom{n}{3}}{\binom{3n}{3}} \times 3$

(b)   Find the probability, $P_s$, that the three cards are all different colours.

It should be one pink card from $n$ pink cards, one yellow card from $n$ yellow cards and one green card from $n$ green cards.
$\therefore P_d = \displaystyle\ddfrac{\binom{n}{1} \times {n\choose 1} \times {n\choose 1}}{\binom{3n}{3}}$

(c)   Find the probability, $P_m$, that the two cards are of one colour and the third is of a different colour.

There are PPY PPG YYP YYG GGP GGY, so six ways.
$\therefore P_d = \displaystyle\ddfrac{\binom{n}{2} \times {n\choose 1}}{\binom{3n}{3}} \times 6$

(d)   If $n$ is large, find the approximate probability ratio $P_s : P_d : P_m$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} &= \ddfrac{\binom{n}{3}}{\binom{3n}{3}} \times 3 : \ddfrac{\binom{n}{1} \times {n\choose 1} \times {n\choose 1}}{\binom{3n}{3}} : \ddfrac{\binom{n}{2} \times {n\choose 1}}{\binom{3n}{3}} \times 6 \\ &= \binom{n}{3} \times 3 : \binom{n}{1} \times {n\choose 1} \times {n\choose 1} : \binom{n}{2} \times {n\choose 1} \times 6 \\ &= \frac{n!}{(n-3)!3!} \times 3 : n \times n \times n : \frac{n!}{(n-2)!2!} \times n \times 6 &\color{green} \binom{n}{k} = \dfrac{n!}{(n-k)!k!} \\ &= \frac{n(n-1)(n-2)(n-3)!}{(n-3)! \times 6} \times 3 : n^3 : \frac{n(n-1)(n-2)!}{(n-2)! \times 2} \times 6n \\ &= \frac{n(n-1)(n-2)}{2} : n^3 : 3n^2(n-1) \\ &= \frac{1}{2}n^3-\frac{3}{2}n^2 + n : n^3 : 3n^3 – 3n^2 \\ &= n^3-3n^2 + n : 2n^3 : 6n^3-6n^2 \\ \text{approximate ratio } &= \lim_{n\to\infty} n^3-3n^2 + n : \lim_{n\to\infty} 2n^3 : \lim_{n\to\infty} 6n^3-6n^2 \\ &= \lim_{n\to\infty} \bigg(\frac{n^3}{n^3}-\frac{3n^2}{n^3} + \frac{n}{n^3}\bigg) : \lim_{n\to\infty} \bigg(\frac{2n^3}{n^3}\bigg) : \lim_{n\to\infty} \bigg(\frac{6n^3}{n^3} – \frac{6n^2}{n^3}\bigg) \\ &= \lim_{n\to\infty} \bigg(1-\frac{3}{n} + \frac{1}{n^2}\bigg) : \lim_{n\to\infty} 2 : \lim_{n\to\infty} \bigg( 6-\frac{6}{n}\bigg) \\ &= 1 : 2 : 6 \end{aligned}

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