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Text Sections: 4.2, 4.3, 5.
Our example from the last lecture: Suppose a coin is slightly bent so that HEADS occurs 55% of the time. You will toss this coin twice. If the tosses are independent, write out the probability for each element of the sample space by filling in the following table:
outcome HH HT TH TT probability
Since our probabilities are independent, we can simply multiply to get results. For instance, the probability of HEADS on First AND HEADS on Second (HH) is just
𝑝𝑟𝑜𝑏 𝐻𝐻 = 0.55 ∙ 0.55 = 0.
Similarly, 𝑝𝑟𝑜𝑏 𝐻𝑇 = 0.55 ∙ 0.45 = 0.
𝑝𝑟𝑜𝑏 𝑇𝐻 = 0.45 ∙ 0.55 = 0.
and 𝑝𝑟𝑜𝑏 𝑇𝑇 = 0.45 ∙ 0.45 = 0.
Filling on our table we have outcome HH HT TH TT probability 0.3025 0.2475 0.2475 0.2025 1.
Note that the probabilities add up to 1 (like they should!). Also, note that we can think about this situation another way as well. Remember that we are really trying to develop a model for surveys with yes/no responses (more generally, polling for a proportion for a dichotomous population). We really only want to know how many in the sample have a certain property. So, for each outcome we can count the number of HEADS. This gives us the following:
Number of HEADS 2 (HH) 1 (HT, TH) 0 (TT) probability 0.3025 0.2475 + 0.2475 = 0.4959 0.2025 1.
See if you can fill in the missing entries in the table for the situation with 3 tosses.
HHH 0. HHT HTH HTT THH 0. THT TTH TTT 1.
And the table below:
Number of HEADS
3 (HHH) 2 (HHT, HTH, or THH)
1 (HTT,THT, or TTH)
0 (TTT)
probability 0.408375 0.091125 1.
So far, so good. We, however, would like to work with much larger numbers. For example, if your sample size (or equivalently your number of coin tosses) is 5 then the number of ways we can obtain, say 3 HEADS is 10. When the sample size is 200, the number of ways to obtain 100 HEADS is 9.054851465610323e+058, where the +58 means multiply by the number 10 a total of 58 times. It’s an astronomical number. To get a sense of how large, try to figure out how many seconds old the universe is. (Use approximately 16 billion years converted to seconds. We’ll need this number later.) Evidently we need to learn how to count.
Simple Counting Arguments :
Remember back to when you were a kid playing with Mr. Potato Head.
Suppose you are putting together a Mr. Potato Head and you have in front of you 4 mouths, 3 noses and 2 sets of eyes. How many different faces can you make?
Multistage Process : Suppose a process will be performed in N stages. For example, N = 4 when we toss a fair coin 4 times. Each of these N = 4 stages can be done in two ways (HEADS or TAILS) so there are 2 ∙ 2 ∙ 2 ∙ 2 = 16 ways to toss a coin (fair or biased) 4 times.
In the Mr. Potato Head example, there are 3 stages: Select a mouth, select a nose, select some eyes. The first stage may be donme in 4 ways, the second in 3 ways and the last in 2 ways.
Here’s an example that is solved every day many times all around the world. We’ll present it in a simplified way. Suppose you wish to visit 5 cities: Albany, Utica, Binghamton, Rochester, and Syracuse. You will visit each one once and only once. How many ways can you do this?
By simply using the multistage process approach we see that we can choose the first city in 5 ways. Say that we start from Utica. Once that’s chosen there are 4 ways to select from the remaining cities (Albany, Binghamton, Rochester, and Syracuse). If you were only visiting 2 out of 5 cities there would be 5 ∙ 4 = 20_._
We’re visiting all 5, so if we choose, say Utica for the first and Rochester for the second then selecting the third from Albany, Binghamton, and Syracuse means there are 5 ∙ 4 ∙ 3 = 60 ways to choose 3 cities. The fourth comes from the remaining 2, then the last is determined for us. All told: 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120 ways to visit 5 cities.
Let’s push this a little further. Suppose you want to visit each state capitol once and only once. There are 50 states, so there are 50 ways to choose the first. If we were only visiting 2 of them there would be 50 ∙ 49 = 2450 ways to plan our trip. 3 cities? 50 ∙ 49 ∙ 48 = 117600. You should be starting to get a little worried. Build a table:
Number of cities (out of the 50 capitols)
Number of ways to plan trip
(^1 ) (^2 50) ∙ 49 = 2450 (^3 50) ∙ 49 ∙ 48 = 117600 (^4 50) ∙ 49 ∙ 48 ∙ 47 = 5527200 (^5 50) ∙ 49 ∙ 48 ∙ 47 ∙ 46 = 254251200 (^6 50) ∙ 49 ∙ 48 ∙ 47 ∙ 46 ∙ 45 ≈ 11441304000
By the time we get to 50 cities there are 3.041409320171338e+064 ways. This number is so large as to be incomprehensible.
Some Notation: Factorials When we multiply numbers like “3 times 2 times 1” we will usually write 3 ∙ 2 ∙ 1 = 3! which is read as “3 factorial”. From the above we have 50! ≈ 3.041409320171338e + 064.
We can probably use some refreshments.
Suppose you have 3 candy bars in front of you, Snickers, a Three Musketeers, and a Baby Ruth. Call them S, T, and B. In how many ways can you arrange these objects in order?
3! = 3 ∙ 2 ∙ 1 = 6
Here they are: STB, SBT, TSB, TBS, BST, BTS.
(Just as a side note: I taught High School some time ago in Cameroon. Some students needed help in combinatorics or counting arguments, (which I had never looked at myself) and I foolishly agreed to help them. I wasn’t able to really “click” with the material until I physically placed objects in front of me on the table top and started manipulating them. You are strongly encouraged to do the same in order to make this all less abstract.)
Suppose now you add a Hershey bar, call it H. How many arrangements are possible now?
4! = 4 ∙ 3 ∙ 2 ∙ 1 = 24
Write them down on a piece of paper!
How many when you add a Reese’s, R?
5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
You can be forgiven if you don’t write these down, but make sure you understand where the numbers are coming from.
Stay with the 5 bars. Suppose you are going to eat three of them, and it matters to you in what order you eat them. In how many ways can this be done?
5 ∙ 4 ∙ 3 = 60
Did you notice that we started out in the form of 5! But lost the last two numbers? We can write:
Permutations :
The number of possible sequences (ordered arrangements) of n objects is
𝑛! = 𝑛 · (𝑛 − 1) · (𝑛 − 2) ·... 3 · 2 · 1
If you are less patient, reason as follows: Within the 60 possible sequences, how many are really just TSH rearranged? Evidently that would be 3 ∙ 2 ∙ 1 = 6 If we “divide these out” we come up with 𝑃5, 3!
These we should all be happy to list by hand: S,T,B,H,R select 3, order unimportant
STB SBH SHR TBH THR BHR STH SBR TBR STR
More formally, the number of possible unordered groups which contain r out of n objects is simply the number of possible permutations, 𝑃𝑛 ,𝑟 divided by the number of ways to arrange r objects (i.e we eliminate the redundant groups). Therefore
Examples
How many (5 card) poker hands are possible if the order of the cards is unimportant?
This is the combination of 52 objects taken 5 at a time or 52! 47!5! = 2598960
In how many distinct ways can you arrange the letters: a, b, c, d?
In how many distinct ways can you arrange the letters: b, o, o, k?
The letters: m, i, s, s, i, s,s, i, p, p, i?
A Last (Hugely Important) Example Suppose you toss a biased coin 6 times. We know that there are 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 = 2^6 = 64 possible
sequences:
HHHHHH HHTHHH HTHHHH HTTHHH THHHHH THTHHH TTHHHH TTTHHH HHHHHT HHTHHT HTHHHT HTTHHT THHHHT THTHHT TTHHHT TTTHHT HHHHTH HHTHTH HTHHTH HTTHTH THHHTH THTHTH TTHHTH TTTHTH HHHHTT HHTHTT HTHHTT HTTHTT THHHTT THTHTT TTHHTT TTTHTT HHHTHH HHTTHH HTHTHH HTTTHH THHTHH THTTHH TTHTHH TTTTHH HHHTHT HHTTHT HTHTHT HTTTHT THHTHT THTTHT TTHTHT TTTTHT HHHTTH HHHTTTH HTHTTH HTTTTH THHTTH THHTTTH TTHTTH TTTTTH HHHTTT HHTTTT HTHTTT HTTTTT THHTTT THTTTT TTHTTT TTTTTT
How many of these have exactly 0 HEADS:
That’s easy, just TTTTTT
How many of these have exactly 1 HEADS:
That’s not too bad, just look through the list to see HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH
How many of these have exactly 2 HEADS:
This is starting to become work. Here’s how we think about it: envision 6 places, like 6 cities, candy bars, etc. Out of these 6, select 2 of them. This is a combination because selecting the third slot for a H and then the fifth gives the same outcome TTHTHT as selecting the fifth and then the third. Evidently we should find
Of these. Take a second and with a highlighter indicate these 15 slots.
How many of these have exactly 3 HEADS:
You should do the same for 4, 5, and 6 HEADS.
