Day 15 Class Notes: Solving Probability Problems by Using Combinations
Suppose we want to find the probability of selecting two picture cards (J, Q, K) when two cards are selected, without
replacement, from a standard deck of cards.
From Day 13 (using the “AND” formula)
P(2 picture cards) = P(first picture card) x P(second picture card) =
Since the “order” of the two picture cards selected is NOT important, this problem can be considered a
“combination probability problem”.
The number of ways 2 picture cards can be selected from 12 possible picture cards is
12
C2 
The number of ways in which two cards can be selected from a deck of 52 is
52
C2 
So the probability of selecting two picture cards is
P(2 picture cards) =
C2

52 C2
12
To get more practice with combinations we will choose to work the probability problems in this section using
Combinations.
Example: A club consists of four men and five women. Three members are to be selected at random to form a
committee. What is the probability that the committee will consist of three women?
Work: The order in which the three members are selected is not important (since we are not attaching labels like
President, Vice-President and Secretary) so we may use combinations.
P(committee consists of three women) = Number of possible committees with three women
Total number of 3 member committees
=
So the probability of the three person committee that is randomly selected from a club of four men and five women
being composed of only women is about
%.
From our “Poker Work” yesterday (Day 14)
Let’s look at the PROBABILITY of being dealt each particular 5 card hand
Note: The number of different 5 card poker hands that could be dealt is 52 C5  2,598,960
Hand
Number of that type of Hand
1 pair
1,098,240
2 pair
123,552
3 of a kind
54,912
Straight
10,164
Flush
5072
Full House
3744
4 of a kind
624
Straight Flush
32
Royal Straight Flush
4
Probability
1,098, 240
 .423  42.3%
2,598,960
123,552
 .048  4.8%
2,598,960
54,912
 .021  2.1%
2,598,960
10,164
 .0039  .39%
2,598,960
5072
 .00195  .195%
2,598,960
3744
 .00144  .144%
2,598,960
624
 .00024  .024%
2,598,960
32
 .0000123  .00123%
2,598,960
4
 .00000154  .000154%
2,598,960
Question: What is the probability of being dealt a 5 card hand that is NONE of the above hands?
Answer: P(none of the hands) = number of 5 card hands that are NONE of those type
total number of 5 card hands
=
or
. That is more than
of the time you will be dealt a hand that is none of those listed above.
Example: Ford is testing 12 new vehicles for possible production. They are testing 3 trucks, 4 minivans and 5
hybrid cars. If we assume that each of the 12 vehicles has the same chance of being selected and 4 new vehicles will
be produced, find the probability that…..
A) no minivans are selected.
P(no mini vans) =
C) 2 trucks and 2 hybrid cars are selected
P(2 trucks and 2 hybrids) =
B) at least one minivan is selected
P(at least one minivan selected) = 1 – P(no minivans)
=