Example 11-3. Given an ordinary deck of 52 cards, suppose it is required
to find the probability of selecting two cards and they are both aces. Here the
probability of selecting an ace on the first draw is P 1 = 4/ 52. If the first card selected
is an ace and it is not put back into the deck, then on the second draw there are only
3 aces left in the deck which now has only 51 cards. Consequently the probability of
getting an ace on the second draw is P 2 = 3 / 51. The required probability of obtaining
an ace on both the first and second drawing of cards is given by the multiplication
principle for independent events so that one can write P=P 1 P 2 =^4
52
·^3
51
=^1
221
The above discussions can be generalized. In studying the occurrence or non-
occurrence of three events E 1 , E 2 , E 3 the probability is denoted
P(E 1 ∩E 2 ∩E 3 ) = P(E 1 )P(E 2 |E 1 )P(E 3 |E 1 ∩E 2 ) (11 .29)
and for independent events
P(E 1 ∩E 2 ∩E 3 ) = P(E 1 )P(E 2 )P(E 3 ) (11 .30)
with similar extensions to a higher number of events taking place.
Example 11-4.
The dice table is completely surrounded with players so that you can see only a
part of the table. A player rolls the dice and you see one die comes up a 6, but you
can’t see the other die. What is the probability the player has rolled a 7 or 11?
Solution: Here there are two events E 1 , E 2 with
E 1 =event one die is a 6
E 2 =event sum of dice is 7 or 11
and we are to find P(E 2 |E 1 ). To solve this problem write down the simple events
as
E 1 ={(6 ,1),(6 ,2),(6 ,3),(6 ,4),(6 ,5),(6 ,6)}
E 2 ={(1 ,6),(2 ,5),(3 ,4),(4 ,3),(5 ,2),(6 ,1),(5 ,6),(6,5)}
with E 1 ∩E 2 ={(6 ,1),(6 ,5)}
Recall the simple events all have equal probabilities of 1 / 36 and consequently
P(E 2 |E 1 ) = P(E^1 ∩E^2 )
P(E 1 )
=^2 /^36
6 / 36
= 1/ 3
Observe that P(E 2 ) = 8 /36 = 2/ 9 =P(E 2 | E 1 ) = 1 / 3. These events are not indepen-
dent. That is, knowing one die is a 6 does effect the probability of the sum being 7
or 11.
