Question

How many ways are there for a horse race with three horses to finish if ties are possible? (Note: Two or three horses may tie.)

Short answer

The required number of ways are\(13\).

Step-by-step solution

Formula of Addition Rule for probability

The probability of two mutually exclusive events is denoted by:\(P(Y{\rm{ or }}Z) = P(Y) + P(Z)\)

The probability of two non-mutually exclusive events is denoted by:\(P(Y{\rm{ or }}Z) = P(Y) + P(Z) - P(Y{\rm{ and }}Z)\)

Use permutation and division rule

First count the number of ways if there are no ties.

In this case any one of the\(3\)horses could finish first and, once you know who is first, the other two horses can finish second and third or third and second. Hence there are\(3 \cdot 2 = 6\)different ways the horses can finish if there are no ties.

There are\(3\)outcomes if exactly\(2\)horses are tied for second. Any one of the\(3\)horses could finish first and then the other two are tied.

Similarly there are\(3\)outcomes with exactly\(2\)horses tied for first.

All that remains is the \(1\) situation where all three horses tie for first. Thus, in total there are \(6 + 3 + 3 + 1 = 13\) different outcomes.