Probality and probality distributions Essay

Submitted By mhsena
Words: 7500
Pages: 30

Chapter 4
Probability and Probability Distributions
Classical Interpretation (Way 1)
This interpretation of probability arose from games of chance. Common sense and theory tell us that:
- The probability of a HEADS from a fair coin is ½

- The probability of an ACE is 4/52 or 1/13

1

- The probability of a SEVEN on two dice is 6/36 or 1/6
First die:
Second die: 1
2
3
4
5
6

1
2
3
4
5
6
7

2
3
4
5
6
7
8

3 4 5 6
4 5 6 7
5 6 7 8
6 7 8 9
7 8 9 10
8 9 10 11
9 10 11 12

We can formally state this as: P (event E) = Ne / N
Where Ne is the outcomes favorable to an event and N is the number of possible outcomes and assuming that all outcomes are equally likely. For the probability of a seven on two dice Ne = 6 and N = 36 so P(7) = 6/36 or 1/6.

2

Relative Frequency Interpretation (Way 2)
This interpretation of probability is an empirical approach to the estimation.
Empirical – (1) Derived from experiment and/or observation rather than theory. (2) Based on experience and observation, rather than systematic logic.
P (event E)  ne / n
Where ne if the number of times event E occurred out of n trials. The real difference between the classical and relative frequency interpretations of probability is that the classical interpretation is based on theory (i.e. known behavior) while the relative frequency interpretation is based on observation.
Finding the Probability of Two Coin Tosses:
First Second
Flip
Flip Outcome
H
HH
H
T
HT
H

TH

T

TT

T

Question: What is the probability of obtaining one head from the two tosses? Order does not matter. What if order does matter?

3

Classical interpretation: As I said earlier, this is based on theory or known mechanism. If the coin is fair then each of the N = 4 outcomes is equally likely. For us, Ne = 2 (TH, HT). Thus,
P(exactly 1 head) = Ne / N= 2 / 4 = ½ or 0.5
Relative frequency interpretation: This requires that we conduct an experiment of n trials where each trial consists of us flipping the coin twice. Let’s assume we did n = 1,000,000 (i.e. a total of
2,000,000 flips which makes 1,000,000 sets of two). I simulated this with SAS. If you want to see the program I have included it on the next page, but it is simply intended for those that have an interest in seeing it. Here is what I got on one of the runs, but do note that you get a different outcome each time if you us a negative value in the Ranuni function call of the SAS program as I have.

Outcome
TT
TH
HT
HH

Frequency (ne)
249,908
250,266
249,565
250,261

Relative frequency
0.249908
0.250266
0.249565
0.250261

P(exactly 1 head)  250,266+249,565 = 0.499831 ≈ 0.5
1,000,000

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data flip; nTT = 0; nTH = 0; nHT = 0; NHH = 0; ntot = 0;
Do While (ntot < 1000000);
* Use of the negative value for the seed results in a different outcome each time because the clock time is used. If you want the same result each time you can use a positive value for the seed in RANUNI; tosscoin1 = Ranuni(-3333); tosscoin2 = Ranuni(-3333); if tosscoin1 < .5 and tosscoin2 < .5 then do; nTT = nTT + 1; end; if tosscoin1 < .5 and tosscoin2 > .5 then do; nTH = nTH + 1; end; if tosscoin1 > .5 and tosscoin2 < .5 then do; nHT = nHT + 1; end; if tosscoin1 > .5 and tosscoin2 > .5 then do; nHH = nHH + 1; end; ntot = ntot + 1; rfTT = nTT/ntot; rfTH = nTH/ntot; rfHT = nHT/ntot; rfHH = nHH/ntot; rf1H = rfTH + rfHT; rf1T = rfTH + rfHT; end; file print; put 'A Simulation of ' ntot 'Sets of 2 Coin Flips.'; put 'In other words, each set of coins is flipped ' ntot 'times'; put 'Number of Observed TT = ' nTT; put 'Relative Frequency of TT = ' rfTT; put 'Number of Observed TH = ' nTH; put 'Relative Frequency of TH = ' rfTH; put 'Number of Observed HT = ' nHT; put 'Relative Frequency of HT = ' rfHT; put 'Number of Observed HH = ' nHH; put 'Relative Frequency of HH = ' rfTT; put 'Relative Frequency of Exactly one Head = ' rf1H; put;
put…