Page 70 - Probability Demystified

CHAPTER 4 The Multiplication Rules 59

The previous example can also be solved using classical probability. Recall
that the sample space for tossing a coin and rolling a die is

H1, H2, H3, H4, H5, H6
T1, T2, T3, T4, T5, T6

Notice that there are 12 outcomes in the sample space and only one
1
outcome is a tail and a 5; hence, P(tail and 5) ¼ .
12

EXAMPLE: An urn contains 2 red balls, 3 green balls, and 5 blue balls.
A ball is selected at random and its color is noted. Then it is replaced and
another ball is selected and its color is noted. Find the probability of each of
these:
a. Selecting 2 blue balls
b. Selecting a blue ball and then a red ball
c. Selecting a green ball and then a blue ball

SOLUTION:
Since the first ball is being replaced before the second ball is selected, the
events are independent.
a. There are 5 blue balls and a total of 10 balls; therefore, the probability
of selecting two blue balls with replacement is
P(blue and blue) ¼ PðblueÞ PðblueÞ
5 5
¼
10 10
25 1
¼ ¼
100 4

b. There are 5 blue balls and 2 red balls, so the probability of selecting a
blue ball and then a red ball with replacement is

Pðblue and redÞ¼ PðblueÞ PðredÞ
5 2
¼
10 10
10 1
¼ ¼
100 10