there are m+n possible ways for either event A or event B to occur,
but only if there are no events in common between them.
n(A or B)=n(A)+n(B)-n(A
B)
Example: Assume you have 20 M&M® brand candies as follows:
5 orange, 6 yellow, 5 red, and 4 green.
In one selection, how many ways can you select either 1 orange or
1 yellow M&M®? What is the corresponding probability?
Answer: Of the 20 M&M's®, 5 are orange and 6 are yellow.
Hence 5+6=11 of the M&M's® are yellow or orange.
The probability of selecting a yellow or orange M&M® is 11/20=0.55.
The M&M's® are either one color or another, hence getting a certain color is mutually exclusive of getting a different color—that is, no M&M's® are rainbow-colored, zebra-striped, or some shade such as orange-yellow or blue-green which thus might be judged different colors by different people. To clarify further the meaning of mutually exclusive, let's say that only one or another event can occur, never both at the same time.
Example: Assume you have 20 M&M's® color distributed as above.
If selected without replacement, in how many ways can you select
two red ones in two selections?
What is the corresponding probability?
Answer: For the first selection, five of the 20 M&M's® are red.
Since we need to get two reds in only two selections, we need only
consider this successful case further, ignoring what happens if
we do not get a red on this first selection.
For the second selection, only four red of the 19 M&M's® remain.
Hence there are 5•4=20 ways of selecting two reds M&M's®
in two selections. The corresponding probability would be:
(5/20)•(4/19)=20/380=1/19 or approximately 0.0526.
The first example above (OR) will be dealt with further below. We will now discuss the second example (AND then). We studied in the last lesson repeated coin flips and die rolls. The size of our sample space, that is the set of all possible outcomes, was the product of the set of possible outcomes for each event: 2•2=4 for two coin flips and 6•6=36 for rolling two dice.
This is often referred to as the Multiplication Rule.
It can only be applied if the events are independent.
For more on that subject see the next lesson.
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If event A can occur in m possible ways
and event B can occur in n possible ways, there are m•n possible ways for both events to occur. n(A and then B)=n(A)×n(B) |
This is generally expressed as event A and then event B occurring. This is an AND situation where both are performed. This calculation extends to three or more events. For example, if event C can occur in o possible ways, there are m•n•o possible ways for these three events to turn out.
Example: How many different ways can parents have three children.
Answer: For each child we will assume there are only two possible
outcomes (thus neglecting effects of extra X or Y chromosomes, or any other
chromosomal/birth defects). The
number of ways can be calculated: 2•2•2 = 8.
These can be listed: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG where
B=boy, G=girl. We could have just as well used the symbols 0 and 1:
000, 001, 010, 011, 100, 101, 110, 111. Note that this is the same
as counting in base 2. This fact can be used to more easily list
outcomes or to check for missing outcomes (exactly 4 have boy first,
exactly 4 have boy second, exactly 4 have boy last, etc).
Another way to represent this information is in tree form with
the branches from each node representing the possibilities for
the next event. Note that this can become very large and thus listing
or displaying the complete sample space is often impractical.
This is often referred to as the Addition Rule.
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If event A can occur in m possible ways
and event B can occur in n possible ways, there are m+n possible ways for either event A or event B to occur, but only if there are no events in common between them. n(A or B)=n(A)+n(B)-n(A B)
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Because often one works with non-overlapping events, you will find that the last term is commonly omitted, but added later. It is better to learn the formula correctly the first time and make a special case when the intersection is indeed empty. An empty intersection might occur due to happenstance or it might occur because the events cannot occur simultaneously, i.e. the events are mutually exclusive. In the M&M® example above, the color selections were mutually exclusive.
We already saw an example of overlapping events when we calculated the probability of the green die having a 2 or the red die of having a 5. A careful inspecation of the diagram in the prior lesson indicates that although there are six outcomes where the green die has a 2 and six outcomes where the red die has a 5, we must be careful not to double count the event where both the green die has a 2 and the red die has a 5. There are thus only 11 not 12 corresponding outcomes and the probability was 11/36 or about 0.306.
Example: Suppose you have four candles you wish to arrange from left to right
on your dinner table. The four candles are vanilla, mulberry,
orange, and raspberry fragrances (shorthand: V, M, O, R).
How many options do you have?
Solution: If you select V first then you still have three options remaining.
If you then pick O, you have two candles to choose from.
You can compute the number of ways to decorate your table by the factoral rule:
for the first choice (event) you have 4 choices;
for the second, 3;
for the third, 2;
and for the last, only 1.
The total ways then to select the four candles are: 4!=4•3•2•1 = 24.
These types of problems occur frequently and can be summarized as follows.
| Factorial Rule: For n different items, there are n! arrangements. |
Another word for arrangements is permutations. More commonly this word is used as in the section below when not all the objects are arranged. Please recall that the symbol ! is mathematical shorthand for factorial. n!=n•(n-1)! and 1!=1. Please also note that by definition and because it makes these types of problems easier, 0!=1. 5! = 5•4•3•2•1 = 120, 4! = 24, 3! = 3•2•1 = 6, and 2!=2.
Try solving this exercise on your own: You need to study, practice football, fix dinner, phone a friend, and go buy a notebook. How many different ways can you arrange your schedule?
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nPr = n! / (n - r)! where r is the number of items arranged from n elements. |
Example:
How many ways can you arrange four figurines from a set of seven?
Answer: 7P4=7!/3!=7•6•5•4=840.
Alternate solution: The figurines can be placed as follows:
7
6 5
4 , which is the same as the factorial notation 7!/3!.
Example: the word MISSISSIPPI contains 11 letters of which 4 are S,
4 are I, and 2 are P. If the S's, I's, and P's are distinguishable there
would be 11! permutations. However, the 4 S's can themselves be arranged
4! different ways as can the 4 I's. The 2 P's have 2!=2 arrangements.
Answer: Thus assuming these repeated elements are truly indistinguishable,
the number of arrangements would be 11!/(4!4!2!)=11!/(4!•4!•2!)=34650.
Example: Consider arranging the letters ABCD. There are 4!=24 such arrangements. If considered as a circular arrangement there are but 3!=6 arrangements.
Often in circular arrangements only betweenness and not clockwise/counterclockwise is what matters. This further reduces the arrangements by a factor of 2.
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nCr = n! / (r!(n - r)!) where r is the number of items taken from n elements. |
Note that these numbers are the same as those in Pascal's Triangle, the binomial formula, and the binomial distribution. Those less than about four digits should become very familiar.
Example: You have five places left for stamps in your stamp book and
you have eight stamps. How many different ways can you select five?
Answer: 8!/(5!3!) = 8•7•6/(3•2)=56.
Think of putting them in slots, the first has eight choices,
the next slot has seven choices and so forth as demonstrated.
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