Probabilities Using Counting Techniques

Factorials

The factorial of a number is the number multiplied by all natural numbers that are lesser in value down to \(1\). A factorial is denoted by the number followed by an exclamation mark (!).

For example, the factorial of \(4\) is:

\(4! = 4 \times 3 \times 2 \times 1 = 24\)

Therefore, we can evaluate the factorial of \(4\) as \(\boldsymbol{24}\).

Permutations

A permutation is an arrangement where the order of the arrangement is important.

To calculate the number of permutations in a set of items is as follows.

Given a set of n items, after choosing r items, the number of possible permutations are:

\(P(n, r) = _nP_r = \cfrac{n!}{(n-r)!}\)

Mark has \(8\) employees, and he has \(5\) different positions that need to be filled that can’t be filled by a different person. How many ways can Mark select and assign his employees to those \(5\) positions?

Here, we are given a set with a size of \(8\) items (or employees) and we need to arrange them in \(5\) positions. In short, \(n=8\) and \(r=5\).

We can plug these values into the Permutations formula and simplify:

\(P(n, r) = _nP_r = \cfrac{n!}{(n-r)!}\)

\(P(8, 5) = \cfrac{8!}{(8-5)!}\)

\(P(8, 5) = \cfrac{8!}{3!}\)

\(P(8, 5) = \cfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}\)

We can simplify this expression even further to make it easier to evaluate. To do so, we can cancel out the \(5!\) that is within \(8!\) as shown:

\(P(8, 5) = \cfrac{8 \times 7 \times 6 \times 5 \times 4 \times \cancel{3 \times 2 \times 1}}{\cancel{3 \times 2 \times 1}}\)

\(P(8, 5) = 8 \times 7 \times 6 \times 5 \times 4\)

\(P(8, 5) = 6720\)

Therefore, out of Mark's \(8\) employees and \(5\) positions, there are \(\boldsymbol{6720}\) ways of arranging them into those different positions.

Combinations

Combinations are similar to permutations in that they involve taking items from a set, the difference between them is that combinations do not take order into account.

To calculate the number of combinations in a set is as follows.Given a set of n items, after choosing r items, the number of possible combinations are:

\(C(n, r) = _nC_r = \cfrac{n!}{(n-r)!r!}\)

Alice has to prepare a test with \(8\) questions for her students and has already thought up \(13\). How many variations of the test can she make?

Here, we are given a set with a size of \(13\) and need to take \(8\) items. In this case, \(n=13\) and \(r=8\).

Next, we can plug these values into the Combinations formula:

\(C(n, r) = _nC_r = \cfrac{n!}{(n-r)!r!}\)

\(C(13, 8) = \cfrac{13!}{(13-8)!8!}\)

\(C(13, 8) = \cfrac{13!}{(5!)!8!}\)

\(C(13, 8) = \cfrac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1) \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\)

Although this is a rather large calculation, we can reduce its size by canceling out both within the numerator and denominator as follows:

\(C(13, 8) = \cfrac{13 \times 12 \times 11 \times 10 \times 9 \times \cancel{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}}{(5 \times 4 \times 3 \times 2 \times 1 \times \cancel{(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}}\)

\(C(13, 8) = \cfrac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1}\)

\(C(13, 8) = \cfrac{154440}{120}\)

\(C(13, 8) = 1287\)

Therefore, we can determine Alice can create \(\boldsymbol{1287}\) possible tests from the questions she came up with.

Probability

Probability as discussed in an earlier lesson, is the chance of an event taking place.

To determine the probability of an event happening, it can be calculated as such:

\(p = \cfrac{\text{Favourable Outcomes}}{\text{Total # of Outcomes}}\)

The value of the probability of an event will always be between \(1\) and \(0\).

NOTE: If the probability is \(1\), then that means the event will always occur. If the probability is \(0\), then that means the event will never occur.

Carrie has an ordinary deck of cards containing \(52\) cards excluding the jokers. The deck contains \(4\) suits (hearts, diamonds, clubs, and spades) with \(13\) cards in each suit. If Carrie picks up \(5\) cards, what is the probability that \(3\) of them are spades?

First, we must know how many combinations of cards there are if you pick up \(5\).

As we can identify \(n = 52\) and \(r=5\), we can plug these values into the Combinations formula and solve:

\(C(n, r) = _nC_r = \cfrac{n!}{(n-r)!r!}\)

\(C(52, 5) = \cfrac{52!}{(52-5)!5!} = \cfrac{52!}{(47)!5!}\)

\(C(52, 5) = \cfrac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1}\)

\(C(52, 5) = \cfrac{311875200}{120} = 2598960\)

Next, we find the favoured combinations. We can divide it into two possible sets of combinations.

First we find how many combinations there are if we choose 3 random cards from the entire set of spades. As we can identify \(n = 13\) and \(r = 3\), we can plug these values into the Combinations formula and solve:

\(C(n, r) = _nC_r = \cfrac{n!}{(n-r)!r!}\)

\(C(13, 3) = \cfrac{13!}{(13-3)!3!} = \cfrac{13!}{(10)!3!}\)

\(C(13, 3) = \cfrac{13 \times 12 \times 11}{3 \times 2 \times 1}\)

\(C(13, 3) = \cfrac{154440}{6} = 286\)

The second set of combinations we must calculate is the opposite of the previous. Because we calculated \(3\) random cards from the set of spades, we must now calculate the combinations from when we choose \(2\) cards from the set of non-spades because that is the favourable outcome. Since we can identify \(n = 39\) and \(r = 2\), we can plug these values into the Combinations formula and solve:

\(C(n, r) = _nC_r = \cfrac{n!}{(n-r)!r!}\)

\(C(39, 2) = \cfrac{39!}{(39-2)!2!} = \cfrac{39!}{37!2!}\)

\(C(39, 2) = \cfrac{39 \times 38}{2 \times 1}\)

\(C(39, 2) = \cfrac{1482}{2} = 741\)

Now we can determine the actual probability by calculating the product of the ideal combinations and dividing that value by the total number of possible combination:

\(p = \cfrac{C(13, 3) \times C(39, 2)}{C(52, 5)}\)

\(p = \cfrac{286 \times 741}{2598960}\)

\(p = \cfrac{211926}{2598960} = 0.08154\)

Therefore, the chances of grabbing \(3\) spades from grabbing a random \(5\) cards is \(\boldsymbol{0.08154}\) or \(\boldsymbol{8.154 \%}\).