Consider an experiment that consists of six horses, numbered 1 through 6, running a race, and suppose that the sample space consists of the 6! possible orders in which the horses finish. Let A be the event that the number-1 horse is among the top three finishers, and let B be the event that the number-2 horse comes in second. How many outcomes are in the event A ∪ B?

Number of outcomes = 432

Step-by-step explanation:

The outcome space will be;

S = { (x1, x2, ... x6), xi∈{1,2, ...6}∀i}

Now, S has all the ordered permutations of the numbers 1-6 which represents horses.

|S| = 6!

So, number of elements in S is 6!

Now, for the event that the number 1 is either x1, x2 or x3;

Let's call the event A;

A = { (1, x2, ... x6), (x1,1, ... x6), (x1, x2,1, ... x6); xi∈{2, ...6},∀i}

So, there are 5! permutations of the other elements of the vector (2,3,4,5,6) and 1 can either be the first, second or third. Thus, by multiplication principle of counting;

|A| = 3 • 5!

Now, let the event where the number 2 is x2 to be B.

Thus;

B = { (x1, 2, ..., x6), xi ∈ {1,3, ...6}∀i}

So, there are 5! permutations of the other elements of the vector (1,3,4,5,6)

Hence, |B| = 5!

Thus, the event A ∩ B where number 1 is the first three and number 2 is second is represented as;

A ∩ B = { (1,2, ... x6), (x1,2,1, ... x6); xi ∈ {3, ...6},∀i}

Now, the remaining numbers, 3,4,5 and 6 can be permitted in 4! ways and either 1 & 2 can be the first and second number or the third and second number.

Thus, |A ∩ B| = 2 • 4!

From set derivations, we know that,

|A ∪ B| = |A| + |B| - |A ∩ B|

So, plugging in the relevant values, we obtain;

|A ∪ B| = (3 • 5!) + (5!) - (2 • 4!)

= (3 x 120) + 120 - (2 x 24) = 432