Q. 2.1 - A First Course in Probability

Question

Consider an experiment that consists of $6$horses, numbered$1$ through$6$, and 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-$1$ the 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$A \cup B$?

Step-by-Step Solution

Answer

$N(A\cup B)=432$

Note that$N(A\cup B)=N\left(A\right)+N\left(B\right)-N(A\cap B)$, and count the elements on the right-hand side.

Step 1 Given Information.

Let$A$ be the event that the number-$1$ the horse is among the top three finishers, and let$B$ be the event that the number-$2$ horse comes in second.

Step 2 Explanation.

Using only mathematical context we have:

Outcome space$S$ :

$S=\left\{\left(x_1,x_2,\dots x_6\right):x_i\in \{1,2,\dots 6\},\forall i\right\}$

$S$has all the ordered permutations of numbers $1-6$(representing horses).

$\left|S\right|=6 !$- the number of elements it $S$is$6 !$.

Event $A$- the number$1$ is either$x_1,x_2$-or$x_3$
.

$A=\left\{\left(1,x_2,\dots x_6\right),\left(x_1,1,\dots x_6\right),\left(x_1,x_2,1,\dots x_6\right):x_i\in \{2,\dots 6\},\forall i\right\}$

There are $5 !$permutations of the other elements of the vector $(2,3,4,5$&$6)$ and $1$can be the first the second, or the third.

By the multiplication principle of counting - $\left|A\right|=3·5!$

Event$B$- the number $2$is$x_2$.

$B=\left\{\left(x_1,2,\dots x_6\right):x_i\in \{1,3,\dots 6\},\forall i\right\}$

There are $5 !$permutations of the other elements of the vector$(1,3,4,5$$\&6)$.$\left|B\right|=5 !$

An event $A\cap B$ where number one is in the first three and the number two is second

$A\cap B=\left\{\left(1,2,\dots x_6\right),\left(x_1,2,1,\dots x_6\right):x_i\in \{3,\dots 6\},\forall i\right\}$

The remaining numbers$-3,4,5$ &$6$ can be permuted in $4 !$ways. And $1,2$can either be the first and the second number or the third and the second number.

$|A\cap B|=2·4!$

Step 3 Explanation.

From one of the previous exercises we know:

$|A\cup B|=\left|A\right|+\left|B\right|-|A\cap B|$

The other notation for the number of elements in a set is$N\left(A\right)$.

Finally:

$|A\cup B|=3·5!+5!-2·4!=18·4!=432$