NOTES LECTURE ON PROBABILITY I
1. Combinatorial Analysis Probability concerns how much possibility an event can occur in a random experiment. For those random experiments with finite number of outcomes, say N , each being equally like, the probability of an event E is defined classically by N (E) , P (E) := N where N (E) denotes the number of outcomes resulting in the occurrence of the event E. Example 1.1. Consider flipping a fair coin, and consider the event that the heads 1 come up. Then N = 2 and N (E) = 1, so the probability of this event is P (E) = 2 . Example 1.2. Consider tossing two dice, and denote by (i, j), i, j = 1, , 6, the outcome where i appears on the first die and j appears on the second die. Let E denote the event consisting of those outcomes (i, j) with i + j = 8. Then 5 N = 6 6 = 36 and N (E) = 5. Thus P (E) = 36 . In order to obtain the probability, we need to calculate the total number of possible outcomes for a random experiment; calculate the number of outcomes of an event in the random experiment. This requires the techniques from combinatorics. 1.1. The counting principle. Proposition 1.1 (The basic counting principle). Suppose that two experiments are to be performed. If the first experiment has m possible outcomes, and the second experiment has n possible outcomes, then there are mn possible outcomes of the two experiments. Proof. The proof follows from enumerating all the possible outcomes. Indeed, suppose the m outcomes of the first experiment are a1 , a2 , , am and the n outcomes of the second experiment are b1 , b2 , , bn . then all the possible outcomes of the two experiments are (a1 , b1 ), (a1 , b2 ), , (a1 , bn ) (a2 , b1 ), (a2 , b2 ), , (a2 , bn ) (am , b1 ), (am , b2 ), , (am , bn ) where (ai , bj ) denotes the outcome that the first experiment has outcome ai and the second has outcome bj . The above table has m rows, and each row has n elements. So there are mn elements in total. The basic counting principle has the following generalization which can be proved by induction.
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Proposition 1.2 (The generalized counting prnciple). Suppose r experiments are to be performed. The experiment 1 has n1 possible outcomes, the experiment 2 has n2 possible outcomes, , and experiment r has nr possible outcomes. then there are n1 n2 nr possible outcomes of the r experiments in total. Example 1.3. The 7-space license plates are designed by occupying the first 3 places by letters and last 4 by numbers. (i) How many different possibilities are there? (ii) If repetition among letters and numbers are prohibited, how many different designs are there? Solution. By the generalized counting principle, we have (i) There are 26 26 26 10 10 10 10 = 175, 760, 000 possibilities. (ii) There are 26 25 24 10 9 8 7 = 78, 624, 000 different designs. 1.2. Permutations. A permutation is an ordered list without repetition. Suppose we have n different objects. Then by the generalized counting principle, there are n (n - 1) (n - 2) 3 2 1(= n!) different permutations of these n objects. Example 1.4. Ten books will be put on a bookshelf. Of these, 4 are math books, 3 are chemistry books, 2 are history books, and 1 is a language book. (i) How many different arrangements are possible? (ii) If the books on same subjects are required to put together, how many different arrangments are there? Solution. (i) There are 10! = 3, 628, 800 different arrangements. (ii) We first divide the bookshelf into 4 parts, there are 4! possibilities to label them as math, chemistry, history and language. Then for each case, there are 4! to ways put math books, 3! ways to put chemistry books, 2! ways to put history books, and 1! way to put language book. By the generalized counting principle, there are 4!4!3!2!1! = 6912 different arragements. Example 1.5. How many different signals, each consisting of 9 flags hung in a line, can be made from a set of 4 white flags, 3 red flags, and 2 blue flags, if all flags of the same color are identical? Solution. If we distiguish the flags with the same color, there are 9! permutations. Since we identify the flags of the same color, different permutations may represent the same signal. By the counting principle, each signal corresponds to 4!3!2! 9! permutations. So the total number of different signals is 4!3!2! = 1260. Remark 1.1. Suppose there are n objects, of which n1 are aloke, n2 are alike, , nr are alike, where n1 + n2 + + nr = n. Then there are n! n1 !n2 ! nr ! different permutations.
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1.3. Combinations. A combination is an unordered colllection of unique elements. Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important. A k-combination is a subset of S with k elements. Let us calculate the number of k-combinations from a set S with n elements. First, assume the order is an issue. Then there are n ways to select item 1, (n-1) ways to select item 2, , (n - k + 1) ways to select item k. So there are n (n - 1) (n - k + 1) ways in total. However, since the order is irrelevant, there are some repetitions. In fact, each group of k elements is counted k! times. So the number of k-combinations is n! n(n - 1) (n - k + 1) = . k! k!(n - k)! Definition 1.1. We will use the notation n! n , = k k!(n - k)! which represents the number of k-combinations selected from n objects. Example 1.6. From a group of 5 women and 7 men, we need to select 2 women and 3 men to form a committee. (i) How many different committee are there? (ii) If 2 of the men refuse to serve together, how many different committee are there? Solution. (i) By the counting principle there are 5 2 7 3 = 350 different
committee. (ii) Since 2 of men refuse to serve together, there are 5 3 selections for men. So there are Proposition 1.3. n k = + 5 2 n-1 k-1 5 2 + 5 2 = 30
30 = 300 different committee. + n-1 k , where 1 k n - 1.
First proof. By definition, we have n-1 k-1 + n-1 k (n - 1)! (n - 1)! + (k - 1)!(n - k)! k!(n - k - 1)! (n - 1)![k + (n - k)] = k!(n - k)! n! = k!(n - k)! n = . k =
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Second proof. Recall that
n represents the number of k-combinations of a set k containing n objects. This number can be calculated in a different way. Let us fix n-1 an object, say the first one, then there are k-combinations that contain k-1 n-1 the first object, and there are also k-combinations that do not contain k n-1 n-1 the first object. So there are + k-combinations in total. k-1 k Theorem 1.1. (Binomial Theorem)
n
(x + y)n =
k=0
n k
xk y n-k .
Proof. We use the induction argument. It is clear for n = 1:
1
k=0
1 k
xk y 1-k =
1 0
x0 y +
1 1
xy 0 = y + x = (x + y)1 .
Suppose it is true for n = l, i.e.
l
(x + y)l =
k=0
l k
xk y l-k .
Then for n = l + 1,
l
(x + y)l+1 = (x + y)(x + y)l = (x + y)
k=0 l
l k l k
xk y l-k xk y l+1-k
=
k=0 l+1
l k
l
xk+1 y l-k +
k=0 l
=
i=1
l i-1
l
xi y l+1-1 +
i=0
l i
xi y l+1-i xi y l+1-i + y l+1
= xl+1 +
i=1 l
l i-1 l+1 i
+
l i
= xl+1 +
i=1 l+1
xi y l+1-i + y l+1
=
i=0
l+1 i
xi y l+1-i .
Thus it is also true for n = l+1. This proves the binomial thereom by induction.
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Homework 1. Problems 1, 7, 9, 13, 17, 18, 20, 24