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STA100 The Binomial Random Variable Lecture 11
Weโre just about done with the โtheoryโ part of the course and are moving into the applications. Most people find that the course gets much more understandable at this point, so ๏ท Good job making it this far! ๏ท Keep moving forward for the sorts of applications you came into the course for.
A Man and his Triangle
You may have noticed a pattern emerging from the previous examples. We would like a formula which tells us, when sampling from a dichotomous population with a sample of size ๐ , what the probabilities
of getting 0, 1, 2 , โฆ ๐ successes are. Think of walking up to individuals in the Mall and asking whether or not they have a match. The response will be Yes or No (Success or Failure) though some
people may also tell you to stop smoking.
๏ท If you sample with sample size ๐ = 1 then your sample space looks like S = ๐, ๐น and you have exactly 1 outcome with 0 successes and exactly 1 outcome with 1 success. This is indicated in the first row of the table. ๏ท If your sample is of size 2 then your sample space looks like S = ๐๐, ๐๐น, ๐น๐, ๐น๐น and you have exactly 1 outcome with 0 successes, 2 outcomes with 1 success, and 1 outcome with 2 successes. This is indicated in the second row of the table. ๏ท One last example here. If your sample is of size 3 then your sample space looks like S = ๐๐๐, ๐๐๐น, ๐๐น๐, ๐๐น๐น, ๐น๐๐, ๐น๐๐น, ๐น๐น๐, ๐น๐น๐น and you have exactly 1 outcome with 0 successes, 3 outcomes with 1 success, 3 outcomes with 2 successes and 1 outcome with 3 successes. ๏ท The following triangle, named after Pascal, tells us the structure of our outcomes for any size sample.
n=1 1 1 n=2 1 2 1 n=3 1 3 3 1 n=4 1 4 6 4 1 n=5 1 5 10 10 5 1 n=6 1 6 15 20 15 6 1 n=
Look at the ๐ = 6 row. The way to construct any row in the table is as follows. First, if you will sample with size 6 that means you will walk up to 6 people and say โDo you have a match?โ There are 22222*2=64 possible outcomes of the form, e.g. SFFFSF (corresponding to first person says Yes, second person says No, third person says No, fourth person says No, fifth
person says Yes, sixth person says No). It gets tedious to list them all out, so we ask โHow many of the 64 possibilities have 0 successes?โ Just 1, of course, the outcome FFFFFF. The we ask, โHow many of the 64 possibilities have 6 successes?โ Just 1, of course, the outcome SSSSSS. Thus the first and last entries in the row must be 1โs, and if you look at the table you will see this is so for any row. How do we do the other entries?
You can think of the number 6 in the second entry of the row labeled โn=6โ as the number of outcomes with exactly 1 success. We know this number is a 6 in two ways. One way is to work the table mechanically, adding the two numbers above our target, to the left and to the right, to get a 6 = 1+5. Then the next entry, corresponding to the number of outcomes with exactly 2 successes is obtained from the table mechanically by adding together the 5 and the 10 to get 15. In this way you can fill out the whole row and then move onto the next row.
But there is another way to think about all of these and, if youโre a little ambitious, to see why the table โworksโ. As we have seen, the number of ways to get 2 successes on 6 trials is the combination of 6 items taken 2 at a time or
How about the entry to the right? The number 20 is (just working the table) the sum of the two entries above to the left and to the right. But also
You should be able to fill in the next row by yourself. Pascalโs triangle helps us to calculate probabilities in our heads for simple sampling situations. Bringing back the ideas from the last lecture (and explained in your book in section 5.2) we have the following.
When an experiment is such that
๏ท you will perform a trial ๐ times, and
๏ท each trial may come up as a success, with probability ๐ , or a failure with probability
๐ โ ๐ (think coin toss or polling dichotomous population) then,
๏ท if each trial is independent of all the others,
๏ท the probability of ๐ outcomes being successful on ๐ trials is
๐๐๐๐ ๐ ๐๐๐๐๐๐๐๐๐ = ๐ช๐,๐ ๐๐^ ๐ โ ๐ ๐โ๐
b. What is the probability that I get at least 4 Hearts in my sample?
_r prob
10._
Since โat least 4โ is the same as โ4 or 5 or โฆ or 10โ we can add the probabilities 0.146+0.058+ 0.016+0.003+0.000+0.000+0.000 to get 0.223.
c. What is the probability that I get at most 3 Hearts in my sample?
_r prob
10._
Since โat most 3โ is the same as โ0 or 1 or 2 o 3โ we can add the probabilities 0.056+0.188+0 .282+0 .250 to get 0.776. Note that these numbers add up to 1.
Some people like to draw a picture instead of a table, so take a look at the following.
Expected Value It turns out that we are interested in the โaverageโ for these experiments
thought of in the following way. Suppose you walk up to 4 people in the Mall and ask โDo you have a match?โ and then write down the number of people who say โyesโ. Your result will be one of 0, 1, 2, 3, or 4. Do this again and again, going from mall to mall. What do you think you will obtain on average? Assuming as before that 15% of people carry matches, I did this 100 times and got the following:
1 0 2 0 0 2 0 1 0 1 0 0 2 0 0 0 0 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1 1 4 0 1 1 1 0 0 0 0 0 0 1 1 2 1 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 1 2 0 1 0 1 1 0 1 1 1 1 0 1 0 0 1
How will we calculate an average?
0 1 2 3 4 5 6 7 8 9 10
0
Probabilities Of r Successes on 10 Trials
r
p
Pretty close. If I repeated the experiment in 1000 different malls Iโll bet Iโd be even closer.
Take a second and, using the table in the appendix, fill in the following for an experiment where you take 6 foul shots (assume your results are independent from shot to shot) and suppose the probability you make any individual shot is 0.7 = 70%.
Number of Shots, r Probability, p 0 1 2 3 4 5 6
total 1
Iโve placed a plot summarizing this situation below.
Suppose you wanted to balance the stems in this figure. Where would you put the fulcrum?
-1 0 1 2 3 4 5 6 7
0
Probabilities for r=number of Successes on n=6 Attempted Foul Shots
Using the table we can calculate the mean, ๐. Hereโs how I would set this up in Excel:
Number of Shots, r Probability, p rp=outcomeprobability 0 .001 0. 1 .010 0. 2 .060 0. 3 .185 0. 4 .324 1. 5 .303 1. 6 .118 0. total 1 4.
Our average, 4.2040, is just about where you would have to balance the plot above.
Summing up
At this point you should feel reasonably comfortable (assuming you are reading the text, working through practice examples from the text, etc.) with ๏ท Binomial Random Variables ๏ท Calculating the average outcome of a random variable.
In the next lecture we learn how to calculate standard deviations and move on to the Normal Distribution.
