Page 211 - Math Course 3 (Book 2)
P. 211
Probability of Compound Events
Dependent Events
Let’s Begin
Example
Independent Events At the school carnival, winners in the ring-toss
game are randomly given a prize from a bag that
Example contains 4 sunglasses, 6 hairbrushes, and 5 key
chains. Three prizes are randomly drawn from the
bag and not replaced.
TRAVEL Find P(sunglasses, hairbrush, key chain).
Rae is flying from Birmingham to Chicago. On the
first leg of her trip, she has to fly from Birmingham
to Houston. In Houston, she changes planes and
heads on to Chicago. The airline reports that the
flight from Birmingham to Houston has a 90% on-
time record, and the flight from Houston to Chica-
go has a 50% on-time record. What is the probabili-
ty that both flights will be on time?
The selection of the first prize affects the selection
of the next prize since there is one less prize from
which to choose. So, the events are dependent.
P(A and B) = P(A) • P(B) Probability of First Prize:
independent 4 ←number of sunglasses
events P(sunglasses) = 15 ←total number of prizes
P(B-H on time = P(B-H on time) • P(H-C on time)
and H-C on Second Prize:
6
time) P(hairbrush) = or 3 ←number of hairbrush
= 0.9 • 0.5 14 7 ←total number of prizes
90% = 0.9 and 50% = 0.5
Third Prize:
5
= 0.45 Multiply. P(key chain) = ←number of key chains
13 ←total number of prizes
The probability that both flights P(sunglasses, hairbrush, key chain) =
Answer P(sunglasses) • P(hairbrush) • P(key chain)
will be on time is 45%.
4 3 5
= • • 13 Substitution
15
7
60 4
= or Multiply
1365 91 Substitution
The probability of drawing
Answer sunglasses a hairbrush, and a
4
key chain is .
91
203