Overview The mathematics of probability underlies a variety of other topics in social science research, including sampling in survey research, significance testing, and maximum likelihood estimation methods.

Contents Key concepts and terms Counting Probability Bibliography




Key Concepts and Terms

1. COUNTING

   • Permutations are how many ways a given number of things can be sequenced. The standard notation is that _nP_n is the permutations (sequences) of all n things in a set of n things; _nP_r is the permutations on n things taken r at a time. The relevant formulas are below:
     _nP_n = n!

     _nP_r = n!/(n - r)!

     where ! is the factorial operation (n! = 1*2*3*...*n). For example, the number of permutations of 4 things taken 3 at a time is 4!/(4-3)! = 24/1 = 24. Thus for items, A, B, C, and D, the 24 permutations are:
     ABC, ACB, ADB, ABD, ADC, ACD
     BAC, BCA, BDA, BAD, BCD, BDC
     CAB, CBA, CBD, CDB, CAD, CDA
     DAC, DCA, DAB, DBA, DBC, DCB
     We could use permutation, for instance, in a study of the effects of ballot order of candidates, when 4 candidates from a possible 10 are to be listed -- we would find there are 5,040 permuations of 10 candidates taken 4 at a time.
     

   • Combinations refer to how many subsets can be derived from a given set, regardless of permutation sequence. For instance, ABC, ACB, BCA, BAC, CAB, and CBA are 6 permutations but only 1 combination. The relevant counting formulas are below:
     _nC_n = 1

     _nC_r = n!/(n - r)!r!

     By cancellation, n things combined 2 at a time = _nC₂ = n!/(n - 2)!2! = (n*(n-1))/2

     For example, the number of combinations of 4 things taken 3 at a time is 4!/[(4-3)!3!] = 24/(1*6) = 4. The 4 combinations of A, B, C, and D are:
     A, B, and C in any order
     A, B, and D in any order
     A, C, and D in any order
     B, C, and D in any order

   • Permutations and Combinations of Sets Not All Different. The foregoing formulas assumed that we were dealing with a set of n things, each of which was different from the other. But what if we have a set of, say, individuals, some of whom are Republicans and some Democrats, and we are interested in permutations by party, not individual? The relevant formulas are below:
     _nC_{n1, n2, ..., nk} = 1

     _nP_{n1, n2, ..., nk} = n!/n1!n2!...nk!

     If we are going go seat all n individuals at a speaker's table and don't care about sequence, there is still only 1 combination of n people taken n at a time. However, if we do care about sequence (permutation), we need to know how many seating arrangements by party affiliation there are of, say, 4 people, 2 of them Republicans and 2 Democrats: ₄P_{2, 2} = 4!/(2!2!) = 24/(2*2) = 6. The six seating sequences are:
     RRDD
     RDRD
     RDDR
     DDRR
     DRDR
     DRRD

     
     

   • Permutations with Recurrences are sometimes needed, as when figuring out how many phone numbers exist in a 7-digit code which, of course, allows any given numeral to recur. The relevant formula is:
     _nP_r = n^r

     Thus, for 10 numerals taken 7 at a time there are 10⁷ = 10 million permutations.

     
     

   • Number of Combinations of Any Size. Sometimes we want to know the number of combinations of any size, as when we might want to know for 6 people, how many committees we could form (that is, of size 6 or size 5 or size 4, etc., down to committess on 1. The relevant formula is:
     _nC_{0, 1,...,n} = 2ⁿ - 1

     The "-1" in the formula above subtracts out the null combination of 0 members. Thus, excluding the null committe, for six people there could be 2⁶ - 1 = 63 possible combinations for committees of any size from 1 to 6.

   
   
   

2. PROBABILITY
   • Notation for probability. Below is a common way by which researchers refer to various types of probability.

     Probability Terms
     P(A)	The probability from 0 to 1.0 that A will occur
     P(AB)	The probability of A or B occurring
     P(AB)	The probability of A and B both occurring.
     P(A|B)	The probability of A occuring, given that B has occurred (conditional probability)

     
     

   • Independent and dependent events. Two events are independent if any given outcome for the first event does not affect the probabilities of any outcome for the second event. Coin tosses are independent events. Two events are dependent if the outcome of the first does affect the second. For instance, the chances of being picked at random to be on a committee is a dependent event because each pick eliminates one person from the pool, changing the probability of being selected for everyone else on the succeeding pick.

   • Probability rules. The rules for probability operations can vary depending on whether one is dealing with independent or dependent events:

     Probability Rules
     	Independent Events	Dependent Events
     P(AB)	P(A)*P(B)	P(A)*P(B|A)
     P(AB)	P(A)+P(B) - P(AB)	same
     P(A|B)	P(A)	(P(AB)/P(B)

     Examples for independent events. For independent events like coin tosses, let P(A) be the probability of getting heads on the first toss, equal to 1/2. Let P(B) be the probability of getting heads on the second toss, also 1/2. The probability of getting two heads in a row (P(AB)) is 1/2 times 1/2 = 1/4. That is, the joint probability of two independent events is the product of their individual probabilities. The probability of getting heads on the first toss or getting heads on the second toss is the sum of their individual probabilities minus their joint probability: P(A)+P(B) - P(AB) = 1/2 + 1/2 -.1/4 = 3/4. The conditional probability is the same as the probability. Thus, the probability of getting heads on the second toss, given that the first toss is heads, is unchanged: P(B|A) = P(B) = 1/2.

     Examples for dependent events. For dependent events like picking committee members from a pool, let P(A) be the probability of picking a Democrat from a pool with two Democrats and two Republicans, so P(A) = 1/2. Let P(B) be the chance of getting a Democrat second, regardless of the first choice. If a Democrat has been selected first, there will be 1 Democrat and 2 Republicans left, giving a conditional probability of picking the Democrat second of 1/3. However if a Republican is picked first, the conditional probability will be 2/3. Overall, the chance is the mean, or P(B) = 1/2, since either condition is equally likely. Now let P(B|A) be the probability that the second person picked is a Democrat, given that the first person picked is a Democrat. Once a Democrat has been picked, there will be 1 Democrat and 2 Republicans left, so the conditional probability of picking a Democrat is 1/3. For dependent events, the joint probability is the probability of the first event times the conditional probability of the second event. Thus P(AB) = P(A)*P(B|A) = 1/2*1/3 = 1/6. That is, there is one chance in six of picking a Democrat first and another Democrat second for the committee, from a pool of 2 Democrats and 2 Republicans. The chance of getting a Democrat first or getting a Democrat second -- P(AB) -- is computed as for independent events and is also 3/4. The conditional probability of picking a Democrat second, given a Democrat has already been picked, P(B|A), equals P(AB)/P(B) = (1/6)/(1/2) = 1/3, as noted above.

   
   




Bibliography

• Mosteller, Frederick, Robert Rourke, and George Thomas (1961). Probability with statistical applications. Reading, MA: Addison-Wesley. A classic text.

Copyright 1998, 2008 by G. David Garson
Last update 3/29/2008.