Chapter 6.3 B
1. Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and
as either domestic or imported. In early 2004, 69% of vehicles sold were light trucks. 78% were
domestic, and 55% were domestic light trucks. Let A be the event that a vehicle is a light truck
and B the event that it is imported. Write each of the following events in terms of events A and
B and give its probability.
Create a Two-Way table to help solve this problem:
Imported
Domestic
Total
Car
.08
.23
0.31
Light Truck
.14
.55
0.69
Total
.22
.78
1
(a) The vehicle is a car 0.31 (Take the complement of P(light truck)
(b) The vehicle is an imported car 0.08
(c) Given that a vehicle is imported, what is the conditional probability that it is a car?
𝑃(𝑐𝑎𝑟|𝑖𝑚𝑝𝑜𝑟𝑡𝑒𝑑) =
.08
.22
= 0.3636
(d) Are the events “vehicle is a car” and “vehicle is imported” independent? Justify your
answer.
The events will only be independent if the probability of one happening is not affected by
the other one happening. For example, P(car) = 0.31. The P(car given we know it is
imported) = 0.3636. The probability of having a car changed if we already know we had an
imported vehicle, so these events can’t be independent.
2. A poker player holds a flush when all 5 cards in the hand belong to the same suite. We will find
the probability of a flush when 5 cards are dealt. Remember that a deck contains 52 cards, 13 of
each suit, and that when the deck is well shuffled, each card dealt is equally likely to be any of
those that remain in the deck.
(a) We will concentrate on spades. What is the probability that the first card dealt is a spade?
What is the conditional probability that the second card is a spade, given that the first is a
space? Remember that this is a conditional probability, and these events are not
independent, so we must use the formula for conditional events: 𝑃(𝐵|𝐴) =
𝑃(𝐴∩𝐵)
𝑃(𝐴)
P(1st card is a spade) = 13/52 = 0.25
P(1st card|2nd is a spade) = P(1st and 2nd spades)/P(1st Spades) = [(13/52)(12/51)]/(13/52) =
12/51 = 0.2353
(b) Continue to count the remaining cards to find the conditional probabilities of a spade on the
third, the fourth, and the fifth card, given in each case that all previous cards are spades.
P(3rd card is a spade| the first two were spades) = [(13/52)(12/51)(11/50)]/0.25 = 11/50 = 0.22
P(4th card is a spade | the first three were spades) = [(13/52)(12/51)(11/50)(10/49)]/.22 = 10/49 = 0.2041
P(5th card is a spade | the first four were spades) = [(13/52)(12/51)(11/50)(10/49)(9/48)]/0.2041 = 9/48 = 0.1875
(c) The probability of being dealt 5 spades is the product of the five probabilities you have
found. Why? What is this probability?
This is the extended multiplication rule. The probability of getting 5 spades in a row is
(13/52)(12/51)(11/50)(10/49)(9/48) = 0.0004952.
(d) The probability of being dealt 5 hearts or 5 diamonds or 5 clubs is the same as the
probability of being dealt 5 spades. What is the probability of being dealt a flush?
Take four times the probability of the event above, because there are four different suites
to get a flush in. 4(.0004952) = 0.001981