Section 12.10: Solving Probability Problems by Using Combinations
To get more practice with combinations we will choose to work the probability problems in this section using
Combinations. The problems we solved in section 12.9 asked the question HOW MANY. We solved the problems
using combinations. No answer in section 12.9 was a fraction. All of the problems in section 12.10 ask the question
– what is the PROBABILITY. We will form fractions to answer each of these questions. All fractions will have the
form
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡ℎ𝑒 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑜𝑢𝑡𝑐𝑜𝑚𝑒 𝑐𝑎𝑛 𝑜𝑐𝑐𝑢𝑟
𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑜𝑢𝑡𝑐𝑜𝑚𝑒 =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑤𝑖𝑡ℎ 𝑛𝑜 𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑡𝑖𝑜𝑛
Both the numerator and the denominator will contain 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
%.
Section 12.10 problem 2: A class consists of 19 girls and 15 boys. If 12 of the students are to be selected at
random, determine the probability they are all girls.
Section 12.10 problem 6: Of 80 people attending a dance 28 have a college degree. If 4 people at the dance are
selected at random, find the probability each has a college degree.
Section 12.10 problem 8: A class of 16 people contains 4 people whose birthday is in October. If 3 people from the
class are selected at random, find the probability that none of those selected has an October birthday.
Section 12.10 problem 12: Duc’s wallet contains 8 bills of the following denominations, four $5 bills, two $10 bills,
one $20 bill and one $50 bill. If he selects two bills at random, determine the probability that he selects 2 $5 bills.
Section 12.10 problem 16: A committee of 4 is to be randomly selected from a group of 7 teachers and 8 students.
Find the probability the committee will consist of 4 students.
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 2 trucks and 2 hybrid cars are selected. This problem is harder than the last
few as it doesn’t ask for all of the objects selected to be the same. The numerator will have to have two
combinations, one for the trucks the other for the hybrid cars. The combinations in the numerator will be
multiplied.
P(2 trucks and 2 hybrid cars) =
Section 12.10 problem 23 – 26 Assume that a particular professional baseball team has 10 pitchers, 6 infielders,
and 9 other players. If 3 players names are to be selected at random, determine the probability that:
23) all 3 are infielders
24) none of the three is a pitcher
25) 2 are pitchers and 1 is an infielder (this is the only one that will absolutely need two combinations in the
numerator)
Section 12.10 problems 27 – 30, a jury pool has 17 men and 22 women, from which 12 will be selected. Assuming
each person is equally likely to be selected and that the jury selected is random, find the probability the jury will
consist of
27) all women
28) 8 women and 4 men
29) 6 men and 6 women
30) I’ll skip this as it is quite a bit more complicated