Tag Archives: P&C

Calculating probability: Any difference whether I use P&C or not?

An example:

We have 10 diamond cards and 3 spade cards whereby each of the cards containing the same shape is assigned to a different number. Find the probability that when I draw 3 cards, there are 2 diamonds and 1 spade cards.

Method 1: Using P&C

Probability required = \displaystyle \frac{{}^{10}{{C}_{2}}\times {}^{3}{{C}_{1}}}{{}^{13}{{C}_{3}}}=\frac{135}{286}

Method 2: Using probability (similar to tree diagram)

Probability required = \displaystyle \frac{10}{13}\times \frac{9}{12}\times \frac{3}{11}\times 3=\frac{135}{286}

A question students may have:

One common question a student may have is: why does Method 2 contain the “\displaystyle \times 3” while Method 1 doesn’t? To investigate, let us apply the definition of combination:

\displaystyle {}^{n}{{C}_{r}}=\frac{n!}{\left( n-r \right)!r!}

For Method 1:

\displaystyle \frac{^{10}{{C}_{2}}{{\times }^{3}}{{C}_{1}}}{^{13}{{C}_{3}}}=\left( \frac{10!}{8!2!} \right)\times \left( \frac{3!}{2!1!} \right)\div \left( \frac{13!}{10!3!} \right)

\displaystyle =\left( \frac{10\times 9}{2} \right)\times \left( \frac{3}{1} \right)\div \left( \frac{13\times 12\times 11}{1\times 2\times 3} \right)

\displaystyle =\frac{10}{13}\times \frac{9}{12}\times \frac{3}{11}\times 3

which gives us the same answer as Method 2.

In Method 2:We are performing the experiment as if we are drawing 1 card at a time.

The reason why we need to multiply by 3 is because to obtain 2 diamonds and 1 spade, the following events are different even though the end product is the same:

a) diamond, followed by diamond, followed by spade

b) diamond, followed by spade, followed by diamond

c) spade, followed by diamond, followed by spade.

However using P & C, for combination, we do not need to consider the order of which card is drawn (or should I say the 3 events are already factored in). You can view it as we are taking 3 cards at one go instead of 1 at a time.