Download How to use Probability Rules to Generate the Numbers of Successes:

How to use Probability Rules to Generate the Numbers of Successes:
Suppose that 20% of people in the USA have blue eyes. Let’s pick a few. Notice that we’re never going to be
picking enough to change the fact that 20% of those remaining in the USA have blue eyes and that 80% have nonblue eyes. In other words, every time we pick, we have a 0.2 probability of someone with blue eyes and an 0.8
probability of someone with non-blue eyes regardless of what we’ve taken away. Each pick is independent of
every other.
1) Pick one person. Let X = the number of blue-eyers we get.
Either you get a blue-eyer (X=1) or you don’t (X=0). The probability of the former is 0.2 and of the latter is 0.8.
X
Probability
0
0.8
1
0.2
2) Pick 2 people. Let X = the number of blue-eyers we get.
b = blue-eyer = probability 0.2 (each time we pick)
n = non-blue-eyer = probability 0.8
X=0
“first is n and second is n” nn
Since they’re independent, we can multiply these probabilities: 0.8 * 0.8 = 0.64
X = 2 bb 0.2 * 0.2 = 0.04
X = 1 nb or bn Since being one sequence means it can’t be the other, we can add these probabilities.
nb: 0.8 * 0.2 = 0.16
bn: 0.2 * 0.8 = 0.16
Adding, we get 0.32
X
Probability
0
0.64
1
0.32
2
0.04
3) Pick 3 people. Let X = the number of blue-eyers we get.
X = 0 nnn 0.8 * 0.8 * 0.8 = 0.512
X = 3 bbb 0.2 * 0.2 * 0.2 = 0.008
X= 1 bnn or nbn or nnb Any of these is 0.2 * 0.8 * 0.8 = 0.128. So our answer is 0.128 + 0.128 + 0.128 = 0.384.
X = 2 nbb or bnb or bbn Any of these is 0.8 * 0.2 * 0.2 = 0.032. So our answer is 0. 032 + 0. 032 + 0. 032 = 0.096.
X
Probability
0
0.512
1
0.384
2
0.096
3
0.008
Note that the probabilities must always add up to 1, so you can double-check.
Our probability histogram is now:
4) Pick 4 people. Let X = the number of blue-eyers we get.
X = 0 nnnn 0.8 * 0.8 * 0.8 * 0.8 = 0.4096
X = 4 bbbb 0.2 * 0.2 * 0.2 * 0.2 = 0.0016
X= 1 bnnn or nbnn or nnbn or nnbn
Any of these is 0.2 * 0.8 * 0.8 * 0.8 = 0.1024.
There are four of them (the single b could be in one of four slots.)
So our answer is 0.1024 * 4 = 0.4096 (coincidentally).
X = 3 nbbb or bnbb or bbnb or bbbn
Notice that you don’t need to write them all out to see that the single n could be in one of four slots.
One way is 0.8 * 0.2 * 0.2 * 0.2. We will be multiplying by 4, so 0.0064 * 4 = 0.0256.
X =2 is more interesting.
(a) We have accounted for 0.4096 + 0.4096 + 0.0256 + 0.0016 = 0.8464.
“X =2” is “NOT (X = 0 or X =1 or X = 3 or X = 4)”. This is to say, the probability of “X =2” is
Everything else: 1 – 0.8464 = 0.1536.
(b) Without getting too deep into “number of ways”, you should be able to see that there are
2 * 2 * 2 * 2 = 16 possible sequences of n’s and b’s. We have used 1 + 4 + 4 + 1 = 10 of them.
Therefore for “two n’s and two b’s” (all that remains), there must be 6 such sequences.
So (for instance) nnbb has probability 0.8 * 0.8 * 0.2 * 0.2 = 0.0256 (coincidentally).
There are 6 sequences like this, so multiply by 6 to get 0.1536.
X
Probability
0
0.4096
1
0.4096
2
0.1536
3
0.0256
4
0.0016
This is as far as we’ll go in computing these by hand (although for a 50/50 split like a fair coin, we might go to n =
5, since we could divide up “what’s left” evenly for X = 2 and X = 3). Without more advanced mathematics, we
can’t really compute these by hand any further without counting “number of ways” of getting particular
sequences.
We can however jump ahead a bit and take a look at n = 80:
Look familiar?