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Unformatted text preview: Chapter 10: Binomial Distribution
A: Combination Number Example 1: From a list of four persons, A, B, C, D a debating team of (a) 1 member (b) 2 members (c) 3 members, is to be chosen. In how many ways can each selection be done? Solution:
(a) Obviously, there are 4 ways of choosing ONE member, from a list of four, to form a team. We use the symbol to represent the number of all possible combinations of “4 choose 1”. ∴ 4 ( )=4
4 1 ()
1 (b) Stage 1: Choose one member from the original four. Stage 2: Choose one member from the remaining three. From the left part, one has, initially, 4 X 3 = 12 counts. However, thinking carefully, with reference to the right part, one sees that actually each team has been counted twice, or that every two counts must be merged into one team. The actual number of teams is ( 4 ) = 4 × 3 = 4 × 3 = 6
2 2 2 ×1 The six teams are: AB, AC, AD, BC, BD, CD. (c)
Stage 1: Choose one member from the original four. Stage Stage 2: Choose one member from the remaining three. Stage 3: Choose one member from the remaining two. Now stages 1 and 2 yield six combinations, AB, AC, BC,…., CD, as explained earlier. For each of these, stage 3 leads to two counts. See the left part of the above diagram. Therefore, the initial number of counts is 6 X 2 =12. However, as the right part of the above diagram shows, each team has been counted three times. Therefore, we must merge every three counts into one team. The actual number of teams is ()
4 3 6 × 2 4 × 3× 2 = = =4 3 3 × 2 ×1 These possible teams are: ABC, ABD, ACD, BCD Example 2: From a list of four persons, A,B,C,D a team of (a) 0 members (a) (b) 4 members is to be chosen. In how many ways can each selection be accomplished? accomplished? Solution: (a) There is only ONE way to do it, namely, choose no person. 4 ∴ ( ) =1
0 (b) There is only ONE way to do it, namely, choose all the 4 members: ABCD. 4 ∴ ( ) =1
4 Factorial Notation
The stepwise multiplication such as 1 x 2 x 3 is a common mathematical calculation. People have used 3! to shorten the writing. It is read as 3 factorial. Thus 3! = 3 x 2 x 1 = 6 In general, r!= r × (r − 1) × ..... × 3 × 2 ×1 From Example 1 and Example 2, we have General case: From n objects we choose r to form a team. The number of possible teams (i.e. combinations) is (i.e. ()
n r n(n − 1).....(n − r + 1) n! = = n Cr = ...........(1) r! (n − r )!r! Example 3: From a list of 49 numbers, we choose six to form a ticket (for the MarkSix lottery). How many tickets are possible? Answer ( )= C = 49! = 49! (49 − 6)!6! 43!6! : 49 × 48 × 47 × 46 × 45 × 44
49 6 49 6 = 6! = 13,983,816 With calculator press: 49 6 = SHIFT
n Cr ÷ 13983816 B: Some rules about ()
n r Example 4: Example 4 From 20 books, we want to choose 17. How many ways are there to do this? Solution: Removing 17 books from 20 is the same as choosing 3 (from 20) to leave behind. Thus, 20 20 ×19 ×18 20 ( 17 ) = ( 3 ) = 3! = 1140 Cr SHIFT Calculator: 20 3 =
n ÷ 1140 The formula n ( ) = ( )......(2)
r n n−r can also be proved using (1), as follows: Multiply both the numerator & denominator of (1) by )........3.2.1 (n − r )(n − r − 1 and we get For n = 0, 1, 2, 3, 4,………….., the various For combinations are: combinations Pascal’s triangle is named after the French Pascal’s mathematician Blaise Pascal (16231662) mathematician However, the triangle was known and studied around 5 centuries earlier by the Chinese mathematician Yang Hui (12381298). But there seems to be a “typo” in the second last row. “34” instead of “35” C: Binomial Probability and Binomial Distn Example 5: You play four games of tennis. For each game, your chance of winning is 0.4 and that of losing is 0.6. Your performance is independent from game to game. What is your probability of winning three games? Solution: Therefore, each favourable case has the same probability, i.e. 3 pq The remaining question is: How many such cases are there? This is the same as asking : This From a list of four boxes, choose three to put ticks on. How many ways are there? How
The answer is: ()
4 3 4 × 3× 2 = =4 3! Hence, the answer to the question: “What is the probability of winning 3 games of tennis?” is: 4 3 3 ( ) p q = 4 × 0 .4
3 × 0.6 = 0.1536 General problem : An experiment is repeated n times. An In each trial, the probability of success is p and In that of failure is q(=1p). q(=1p). The trials are independently performed. The What is the probability of getting r successes? Answer: Generalizing from the above idea, we have: n r n for r=0,1,…,n − r r P( X = r ) = ( )p q Now, the above concerns experiments with just two (i.e. bi ) outcomes, success or failure. Therefore, it is called a binomial probability. Example 6: Find the probability of obtaining r successes in n=4 trials where r =0,1,2,3,4. =0,1,2,3,4. Notes: (i) For P(3) above, the steps on the calculator are: ∧ ON 4 x 0.4 3 ) x 0.6 EXE 0.1536=P(3) In the above, the full list of the five binomial In full probabilities for the case of n=4 and p=0.4 is called a binomial distribution. binomial There are many binomial distributions. Table 2 (to be shown next time) gives these distributions for n=1,2,…..,14 and for p=0.05,0.10,…..0.50. p=0.05,0.10,…..0.50. To prove/check that Note (ii) p = 0 .4 = 1
0 0
1 ∧ Press: ON 0.4 0 EXE Example 7: Find 8! and ( )
12 5 x! n Cr Note (iii) 50FH has and keys. SHIFT Solution: ON 8 EXE x! x −1 40320 SHIFT n Cr AC 12 5 EXE 792 Note (iv):
The combination numbers ()
n r are also known as the Binomial coefficients, as they appear in the binomial expansion: ( x + y) =
n ( )x + ( )x
n 0 n n 1 n −1 y + ...... + ( )x
n r n− r y + ...... +
r ( )y
n n n Simple Exercise
Find 10! and ()
30 25 ...
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 Spring '11
 Gabrille
 Binomial

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