Example 11-5.
Two cards are selected at random from an ordinary deck of 52 cards. Find the
probability that both cards are spades. Find the probability that the first card is a
spade and the second card is a heart. In performing this experiment assume that
the first card selected is not replaced in the deck.
Solution: Examine the events
E 1 =The event that the first card selected is a spade.
E 2 =The event that the second card selected is a spade.
E 3 =The event that the second card selected is a heart.
Using elementary probability theory
P(E 1 ) =^13
52
= Number of spades in deck
Total number of cards in deck
Now if event E 1 has occurred, the deck now has only 51 cards with 12 spades.
Consequently, the conditional probability is
P(E 2 |E 1 ) =
12
51 =
Number of spades in deck
Total number of cards in deck
and P(E 3 |E 1 ) =
13
51 =
Number of hearts in deck
Total number of cards in deck
Calculate the probabilities
P(E 1 ∩E 2 ) = P(E 1 )P(E 2 |E 1 ) =^13
52
·^12
51
=^3
51
and P(E 1 ∩E 3 ) = P(E 1 )P(E 3 |E 1 ) =
13
52 ·
13
51 =
13
204
Permutations
Assume something can be done in different ways and one of these -ways has
been done. Now if a second something can be done in mdifferent ways, then the
number of ways that the two somethings can be done is given by the product ·m.
If a third something can be done in nways, then the three somethings can be done
in ·m·n-ways. This multiplication principle can be extended if more than three
somethings are involved in the study.
