EX 2.16 equally likely outcomes

You have

• six unread mysteries
• six unread science fiction books on your bookshelf. And
• The first three of each type are hardcover
• the last three are paperback.

Consider

• randomly selecting one of the six mysteries

• then randomly selecting one of the six science fiction books

• to take on a post-finals vacation to Acapulco

  • after all, you need something to read on the beach

• Number the mysteries $1,2,\dots ,6$ ,

  • do the same for the science fiction books.

• Then each outcome is a pair of numbers

  • such as $(4,1)$,

• there are $N=36$ possible outcomes

  • For a visual of this situation, refer to the table in Example 2.3 and delete the first row and column

With random selection as described, the 36 outcomes are equally likely. Nine of these outcomes are such that both selected books are paperbacks

• (those in the lower right-hand corner of the referenced table):
• $\left(4,4\right)$ , $\left(4,5\right)$, …, $\left(6,6\right)$. So the probability of the event $A$ that both selected books are paperbacks is

$$P\left(A\right)=\frac{N\left(A\right)}{N}=\frac{9}{36}=.25$$

Question

Is there any other way to obtain $P(A)$?