Exam Advice
Counting problems can be extremely complex and difficult. However, previous SOA exam counting problems have been relatively simple and straightforward.
Focus on the basics. Don’t spend too much time trying to master difficult problems.
Focus on the essence. Don’t attempt to memorize the precise definition of terms such as
“sampling with (or without) order and with (or without) replacement.” As long as you know how to calculate the number of selections, it is not necessary to know whether a sampling method is with (or without) order and with (or without) replacement.
Multiplication Rule
You are going from City A to City C via City B. There are m different ways from City A to City B. There are n different ways from City B to City C. How many different ways are there from City A to City C?
There are m n× different ways from City A to City C.
A B C
m n
Example. A department of a company has 3 managers, 10 professionals, and 5
secretaries. You want to choose one manager, one professional, and one secretary to form a committee. How many different ways can you form a committee?
Solution
There are 3 ways to choose a manager, 10 ways to choose a professional and 5 ways to choose a secretary. As a result, there are 3 10 5 150× × = different ways to form a committee.
Addition Rule
If Event A and B are mutually exclusive, then Pr
(
A B)
=Pr( )
A +Pr( )
BFor any two events A and B , Pr
(
A B)
=Pr( )
A +Pr( )
B Pr(
A B)
If you understand the above formulas, then you have implicitly understood the addition rule without explicitly memorizing it.
Basic terms – Understand the essence. No need to memorize the definition.
Sample with order – The order by which items are listed matters. For example, two of the five equally qualified candidates (A B C D E, , , , ) are chosen to fill two positions in a company --Vice President and Sales Manager. If we first list the candidate chosen for the Vice President position and then the candidate for the Sales Manager position, we can assume the order by which candidates are listed is important. Different orders represent different choices. For example, AB is different fromBA. AB andBA two distinct entities.
Sample without order – The order by which items are listed does not matter. For
example, two of the five equally qualified candidates (A B C D E, , , , ) are chosen to form a committee. Evidently, a committee consisting ofAB is the same committee consisting of
BA.AB and BA are an identical entity.
Sample with replacement – After an item is taken out from the pool, it is immediately put back to the pool before the next random draw. As a result, the same item can be selected again and again. In addition, the number of items in the pool stays constant before each draw.
For example, you have 3 awards to give individually to 5 potential recipients. If one recipient is allowed to get multiple awards, then this is a sample with replacement.
Sample without replacement – Once an item is taken out of a pool, it permanently leaves the pool and can never be selected again. The number of items in the pool always
decreases by one at the end of each random draw.
You have 3 awards to give individually to 5 potential recipients. If no recipients are allowed to get more than one award, then this is a sample without replacement.
Of course, you can sample with (or without) order and with (or without) replacement.
These terms are explained below. Once again, when studying these terms, focus on learning the calculation, not the precise definition of the terms.
Select r out of n people to form a line (or select r out of n distinct objects) -- order matters and duplication is NOT allowed (or duplication is NOT possible) (called “sampling with order without replacement”)
How many different ways can you form the line?
You can start filling the empty spots from left to right (you can also form the line from right to left and get the same result). You have n choices to fill the first spot,
(n 1)choices to fill the second spot,…, and (n r+1) choices to fill the r-th spot.
Applying the multiplication rule, you have a total of
( 1) ( 2) ... ( 1) nr
n× n × n × × n r+ =P ways of forming the line.
If r= , then we have n n!ways of forming the line.
Line up n= + + +r1 r2 ... rk colored balls where r balls have color 1 (such as red), 1 r2 balls have color 2 (such as black), …, r balls have color k k
How many distinct ways are there to put n balls in a row? We have a total of n! permutations if each ball is distinct. However, we have identical balls and we need to remove the permutations by these identical balls. So r1 balls are identical and have r1! permutations; r2 balls are identical and have r2!permutations. …rk balls are identical and have rk!permutations. Thus, we need to divide n! by r r1! !... !2 rk to get the distinct number of permutations.
The total number of distinct permutations is:
1 2 many distinct symbols can you create?
Solution
Select r out of n (where n r) people to form a committee – order does NOT matter and duplication is NOT allowed (called “sampling without order and without replacement”)
How many different ways can you form a committee?
First, you can choose rout of n people to form a line. You have a total of
( 1) ( 2) ... ( 1) nr
n× n × n × × n r+ =P ways to form a line with exactly rpeople standing in line. Out of these Pnr ways, the same committee is repeated r!times. As a result, you have a total of
!
! !( )!
r
n r
n
P n
r =r n r =C
ways of forming a different committee.
r
Cn is often called the binomial coefficient.
If you have difficulty understanding why we need to divide Pnrby r!, use a simple example. Say you are selecting two members out of three people A B C, , to form a committee. First, you choose 2 out of 3 people to stand in a line. You have a total of
2
3 3 2 6
P = × = ways of forming a 2-person line:
, , , , ,
AB BA AC CA BC CB
Of these 6 ways, you notice that the committee consisting of ABis the same as the committee consisting of BA; ACis the same as CA; BCis the same as CB. In other words, the number of committees is double counted. As a result, we need to divide P32 by 2! 2= to get the correct number of committees that can be formed.
Form r-lettered symbols using n letters -- order matters and duplication is allowed (called “sampling with order and with replacement”)
Example. Out of five letters a b c d e, , , , , you are forming three-lettered symbols such as aaa, aab,abc, …. How many different symbols can you build?
You can use any of the n letters as the first letter of your symbol. You can use any of the n letters as the second letter of your symbol (because duplication is allowed). … You can use any of the n letters as the r-th letter of your symbol.
Applying the multiplication rule, we find that the number of unique symbols we can create is:
... r r
n n× × n=n
Example. You are using four letters a b c d, , , to form a three-letter-symbol (for example, adb is one symbol). You are allowed to use the same letter multiple times when building a symbol. How many symbols can you form?
Solution
You can form a total of 43 =64 different symbols.
Collect a total of n items from rcategories – order doesn’t matter and duplication is allowed (called “sampling without order and with replacement”)
This is best explained using an example. You want to collect a dozen silverware items from 3 categories: spoons, forks, and knives. You are not obligated to collect any items from any specific category, but you must collect a total of 12 items.
You have many choices. You can have 12 spoons, zero forks and zero knives; you can have 5 spoons, 5 forks, and 2 knives; you can have zero spoons, zero forks, and 12 knives. The only stipulation is that you must collect a total of 12 items.
How many different collections can you have?
To solve this problem, we need to use a diagram. We have a total of 14 empty boxes to fill. These 14 boxes consist of 12 silverware items we want to collect and 2 other boxes that indicate the ending of our selection of two categories.
For example, we want to collect 5 spoons, 4 forks, and 3 knives. From left to right, we fill 5 empty boxes with spoons. In the sixth box, we put in an “X” sign indicating the end of our spoon collection. We then start filling the next 4 empty boxes with forks. Similarly, at the end of our fork collection, we fill an empty box with the “X” sign indicating the end
of our folk collection. Finally, we fill the 3 remaining empty boxes with knives.
However, this time we don’t need any “X” sign to signal the end of our knife collection (our collection automatically ends there).
Because we have a total of 3 categories, we need to have 3-1=2 “X” signs.
This is why we have a total of 14 boxes. See diagram below.
Similarly, if we want to collect zero spoons, 8 forks, and 4 knives, the diagram will look like this:
If we want to collect zero spoons, zero forks, and 12 knives, the diagram will look like this:
Now we see that by positioning the two “X” boxes differently, we have different collections. We can put our two “X” boxes anywhere in the 14 available spots and each unique positioning represents one unique collection. Because we have a total of C142 different ways of positioning the two “X” boxes, we have a total of C142 different collections.
Generally, we have Cn rr+1 1=Cn rn+ 1 different ways to assemble n items from rcategories.
Example. Assume that a grocery store allows you to spend $5 to purchase a total of 10 items from a selection of four different vegetables: cucumbers, carrots, peppers, and tomatoes. Of these four categories, you can choose whatever items you want, but the total number of items you can have is 10. How many different vegetable combinations can you get with $5?
Solution
X X K K K K K K K K K K K K
X F F F F F F F F X K K K K
S S S S S X F F F F X K K K
You have 286 ways of buying 10 items.
SOA Problem (#1 Nov 2001)
An urn contains 10 balls; 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.44. Calculate the number of blue balls in the second urn.
( ) 4A ( ) 20B ( ) 24C ( ) 44D ( ) 64E Solution
Let x =# of blue balls in the second urn.
1st Urn 2nd urn
Red ball 4 16
Blue ball 6 x
Total 10 x+16
Pr(red ball) 4/10 16/(16+x)
Pr(blue ball) 6/10 x/(16+x)
Pr(both red) + Pr(both blue)=0.44
Because the first urn and second urn are independent, Pr(both red)= 4 16
10 16× x
+ , Pr(both blue)= 6 10 16
x
× x + So we have
4 16 6
0.44 4
10 16 10 16
x x
x x
× + × = =
+ +
Homework for you – redo all the problems listed in this chapter.