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ch11applicationsofbinomialprobabilities - Chapter 11...

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Unformatted text preview: Chapter 11: Chapter Applications of Binomial Probabilities Applications For repeated, independent, experiments For with just two outcomes, success and failure, where P (success) = p and failure, P (failure) = q = 1-p, the probability of 1-p the gaining x successes out of n trials is trials P( X = x) = ( ) p q n x x n− x , x = 0,1,....., n The P(X=x) is called a binomial probability, and the whole list of P(x) for x=0,1,2,….,n is called a binomial distribution. We use the symbols X ~ Bin (n, p) Bin to stand for the sentence, to “X is distributed like a binomial model with number of trials n and probability of success per trial p” success The values of the binomial probabilities can be computed directly. They are also tabulated for certain selected values of n and p in Table 2 which is an extract from a much larger book for binomial distributions. much A: Straightforward Applications Example 1(a): A plays B in the final of a badminton tournament and each performs independently from game to game. Whoever wins two games out of three first gets the gold medal. In each game, A’s chance of winning is 0.45 while B’s chance of winning is 0.55. What is the probability that A gets the medal? What Solution: Solution Assume that they play all the three games, Assume whether or not the match ends before the 3rd whether game. game. Let X = the number of games A wins in these three three Then X ~ Bin( n = 3, p = 0.45) From Table 2, one finds, for n=3 and p=0.45, From P(2)=0.3341 and P(3)=0.0911 P(2)=0.3341 Therefore, Therefore, P(A gets gold medal) P(A =P(X≥2) =P(X=2)+P(X=3)= 0.3342 + 0.0911 =0.4252 Example 1(b) A plays B in the final of a badminton tournament. Who ever wins three games first gets the cup. In each game, A’s chance of winning is 0.65. Both perform independently from game to game. What is the probability that A gets the cup? What Recommended Method: X = the number of times A wins X ~ Bin ( n=5, p=0.65) P(A gets the cup)=P(X=3) + P(X=4) + P(X=5) look up Binomial tables = 0.3364 + 0.3124 + 0.1160 = 0.7648 This method is recommended by Prof Chiu and Gabrielle! Let’s look at Example 1(b) in more detail: The way of working out the answer on this page is a method that many students like to do. This method works BUT is NOT recommended by Gabrielle or Prof Chiu because it is not efficient (It’s correct but NOT fast!) Who ever wins 3 games first! Possibilities: AA | A A=3 BAA | A ABA | A AAB | A BBAA | A BABA | A BAAB | A ABBA | A ABAB | A AABB | A Method B: P(A gets the cup) B=0 A=3 B=1 A=3 = B=2 ( )(0.65) (0.65) + ( )(0.65) (0.35)(0.65) + ( )(0.65) (0.35) (0.65) 2 2 2 3 2 2 4 2 2 2 = (0.65) 2 (0.65) + 3(0.65) 2 (0.35)(0.65) + 6(0.65) 2 (0.35) 2 (0.65) = (0.65)3[1 + 1.05 + 0.735] = 0.7648 Let’s look at a much BETTER (Prof Chiu’s) method: Who ever wins 3 games first! Possibilities: AA | A A=3 B=0 BAA | A A=3 AAA BB AAA AA AAA BA AAA AB B=1 BAA AB BAA AA ABA | A A=3 B=1 ABA AA ABA AB AAB | A A=3 B=1 AAB A A AAB AB And so on……………………………………………… Think of this like the MEN’S Tennis finals in Melbourne or Wimbledon Recommended Method: X = the number of times A wins X ~ Bin ( n=5, p=0.65) P(A gets the cup)=P(X=3) + P(X=4) + P(X=5) look up Binomial tables This method is recommended and is MUCH FASTER than the other way Example 2: Example The probability that a soldier returns from a battle uninjured is 0.85. A group of five is sent out. What is the probability that at least four of them will return uninjured? them (Assume the independence of probabilities) Solution: Solution: Let s = number of soldiers who return uninjured. Then s ~ Bin (n=5, p=0.85) Then Table 2 does not give binomial probabilities for p more than 0.5. We can make use of the block for n=5 and n=5 p=0.15, p=0.15 but downside up. downside Thus, P(s≥4) Thus, =P(4)+P(5) =0.3915 +0.4437 =0.8352 =0.8352 Note : “At least 4” means 4 or 5 soldiers returning UNINJURED. UNINJURED. Thus, an alternative method is to look at Thus, r, which represents the number of returning which INJURED and for which, r ~ Bin (n=5, p=0.15) Bin Therefore, P (r ≤ 1) = P(0) + P(1) Therefore, = 0.4437 + 0.3915 = 0.8352 0.8352 Example 3 Example A large batch of apples is delivered to a customer, who adopts a twostage sampling inspection. Stage 1: He takes a sample of 10 apples. He If NO defective apple is found, he accepts the batch. If two or more defectives are found, he rejects the batch. If If only one defective is found, he takes a 2nd sample, i.e. he goes to stage 2. stage Stage 2: He takes another sample of 14 apples. He If 0 or 1 defective apple is found, he accepts the batch. Otherwise, he rejects the batch. Otherwise, Suppose the proportion of the batch that is actually defective is 0.15. Suppose What is the probability that it will be accepted? What Draw a diagram to summarize this question Draw Solution: Solution Stage 1: Let X = the number of defectives out of 10 Let then X ~ Bin (n=10, p=0.15) P (batch Accepted at stage 1) (batch = P(X=0) = 0.1969 from Table 2 0.1969 P(2nd sample needed) = P(X=1) = 0.3474 from Table 2 0.3474 Stage 2: Stage Let Y = the number of defectives out of 14. Let then Y ~ Bin (n=14, p=0.15) P (Accepts batch at stage 2) = P(Y=0) + P(Y=1) P(Y=0) = 0.1028 + 0.2539 = 0.3567 from Table 2 0.3567 Combining the analyzed results of Stages 1 and 2, we have Stages P (customer accepts batch) (customer = P(X=0) + P(X=1)[P(Y=0)+P(Y=1)] P(X=0) =0.1969 + (0.3474 x 0.3567) =0.3208 Note: Acceptance sampling is an important topic Acceptance in Quality Control. in Example 4 Example Draw the graph for each of the following Draw distributions: distributions (a) Bin(6,0.1) (b) Bin(6,0.5) (c) Bin(6,0.85) From Table 2, we have these figures from From which bar charts and histograms can be drawn: drawn: B: The Mean and Variance of a Binomial Distribution Distribution Whenever one is given a set (i.e. a distribution) of data, Whenever whether it is presented in a frequency table like Chapter 5 Table 6, or in a proportion table like Chapter 6 Example 1, one’s job is to summarize information into two quantities, mean, μ ,and variance, σ² (or equivalently, standard deviation, σ) Now we are studying binomial distributions. We have, for each distribution, a proportion (i.e. probability) table. Our next task would be to find its mean and variance. Our Example 5: Example For the distribution X ~ Bin (n=4,p=0.4), find For the mean and variance. the Note: Note: Solution: Here, we have scores and Table 2 gives probabilities so we Bin (n=4, p=0.4) distn: Bin should use the formulas from Chapter 6 from µ = ∑xi pi σ = ∑ x . pi − µ 2 2 i 2 With calculator: With ON, MODE, COMP, MODE, SD 0 SHIFT ; 0.1296 DATA 1 SHIFT ; 0.3456 DATA ……….. 4 SHIFT ; 0.0256 DATA SHIFT S-VAR 1.6 = µ x SHIFT S-VAR xσ n x 2 0.96 = σ 2 The above results show that for X ~ Bin (n, p) distribution: µ X = np, Bin σ = npq 2 X In fact, they are always true. In You can verify these on every block of figures in Table 2. In fact, the above equations can be proven mathematically, although we prefer to omit the proof here. These formulas are very convenient, for we do NOT need to perform detailed computations every time. every Example 6: Example It is known that a binomial distribution has It mean 36 and variance 19.8 What are its two parameters, n and p? parameters Solution: µ = np = 36 σ 2 = npq = 19.8 npq 19.8 ∴ = np 36 q = 0.55 ∴ p = 1 − q = 1 − 0.55 = 0.45 36 36 n= = = 80 p 0.45 More examples (a bit tricky) More A student takes an exam in which there are student 5 hard questions and 5 easy questions. The student answers the questions independently. Let X = the number of questions that the student answers correctly. Does X have a binomial distribution? Does Solution: Solution: Let X=the number of correctly answered questions, in n=10 Let trials. Questions are answered independently Questions What about the value of p? The wording of this question suggests that the probabilities The of answering each question correctly are NOT the same, presumably being smaller for hard questions than for easier questions. than Therefore, X does NOT have a binomial distribution. Therefore, NOT Another example: Another Players A and B are evenly matched in ability for Players playing handball. They agree to continue to play games of handball until player A wins a game. Each game is played independently of all other games. games. Let Y be the number of games they play. Let Does Y have a binomial distribution? Does Solution: Solution: Since A and B are said to be evenly matched, we can Since assume that the probability that A wins any given game is p=0.5. Also, games are played independently. However, the number of trials is NOT fixed the and we are NOT counting successes. NOT In fact, the random variable Y is counting trials, NOT successes. Thus, Y does NOT have a binomial distribution. NOT binomial The distribution of Y is actually GEOMETRIC with parameter p=0.5 which we are NOT studying in this course. which A very fishy example: very A pond is stocked with 100 fish of which 40 are pond of legal size. You catch 3 fish. Let X=the number of fish of legal size that you catch. catch. Does X have a binomial distribution? Answer: Maybe! It depends! Answer It matters whether you put back the first fish you catch It OR you decide to take home and eat the first fish you You can think of the process of catching fish as if it were a You random sample of n=3 fish taken from the pond. Each fish caught is a Bernoulli trial, with “success” being when the fish is of legal size. when Note that a fish has probability p=of being legal size. The only fact that remains is whether the trials (fish) The are independent. IF each fish is RETURNED to the pond after it is caught each and a caught fish fails to learn from its experience, the fish are being sampled WITH REPLACEMENT. fish In this case the trials are independent and X has a independent binomial distribution with parameters n=3 and p=0.4. binomial We write X ~ Bin ( n=3, p=0.4 ) We Fishy example cont’d Fishy On the other hand, most fishermen do NOT return caught fish to the pond; in On NOT this case, the fish are sampled WITHOUT REPLACEMENT. WITHOUT If the first 2 fish you catch are of legal size, the third fish is sampled from a population of 98 fish of whom 38 are of legal size. Hence, 38 P(3rd fish is legal | first 2 fish legal ) = = 0.388 ≠ 0.4 = P(3rd fish is legal size) P(3rd 98 So the trials (fish) are DEPENDENT. So DEPENDENT Thus, X does NOT have a binomial distribution. NOT binomial It has a HYPERGEOMETRIC distribution. It HYPERGEOMETRIC It’s really Fishy It’s It has a HYPERGEOMETRIC distribution. It HYPERGEOMETRIC Let S=”success”= fish is legal size Let =”success”= legal F=”failure”= fish is NOT legal size =”failure”= NOT All the possibilities of catching 3 fish are: SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF We want only 2 successes, so the possibilities are: We successes P ( X = 2) SSF, SFS, FSS = ( 40 x 39 x 60 ) + ( 40 x 60 x 39 ) + ( 60 x 40 x 39 ) 100 99 98 = 3 x0.0965 = 0.2895 100 99 98 100 99 98 or use the hypergeometric formula: C (40,2) xC (60,1) C (100,3) 40! 60! x 38!2! 59!1! = 100! 97!3! = 0.2895 P ( X = 2) = Past final exam questions on Binomial Distribution Binomial Articles are produced by independent operations of a Articles machine. At any one operation, the probability is 0.8 that the article is acceptable. The articles are placed in boxes as they are produced. The Each box contains 5 articles. Each (a) What is the probability that a randomly selected box contains no acceptable articles? contains (b) What is the probability that a randomly selected box (b) contains more acceptable than defective articles? contains (c) Of 20 boxes examined, what is the probability that at most (c) one contains no acceptable articles? one Solution: Let X = number of acceptable articles in each box (the number of non-defective articles) X~ Bin (n=5, p=0.8) Use the binomial formula : (a) P[X=x] = ( )p n x x (1 − p) n − x P [Box contains no acceptable articles] = P[X=0] = 0.00032 You can use your calculator BUT it’s QUICKER to use table 2: Look up n = 5 , p = 0.2 , x = 5 because “no acceptable articles in each box” is the same as “all 5 articles in each box are DEFECTIVE” The probability that the articles are defective equals 1- 0.8 = 0.2 Solution: (b) Solution: P [Box contains more acceptable than [Box defective articles] = P[X=3] + P[X=4] + P[X=5] P[X=3] = 0.2048 + 0.4096 + 0.32768 0.2048 = 0.94208 Using Table 2 is much quicker than the calculator! calculator! Solution: (c) Solution: Let Y = the number of boxes containing no boxes acceptable articles acceptable Y ~ Bin ( n=20, p = 0.00032) from (a) Bin P [Of 20 boxes examined , at most one contains no acceptable articles] articles] = P[Y=0] + P[Y=1] P[Y=0] = 0.99362 + 0.006361 = 0.99998 0.99998 You CANNOT USE table 2 for this question. You CANNOT IT MUST be done with your calculator, so don’t forget to practise with the 50FH! BINOMIAL and NORMAL Distributions in the same question the The serum cholesterol level X in 14-yr-old boys has The approximately a normal distribution with mean µ= 170 and 170 variance =900. variance (a) Find the probability that a randomly selected 14-yr-old boy (a) has serum cholesterol level between 160 and 200. has (b) Find the probability that exactly 7 boys out of 10 randomly (b) selected 14-yr-old boys, have serum cholesterol levels between 160 and 200. between (c) Usually high serum cholesterol levels are a cause for (c) concern. Find the level c such that 1% of the population of 14-yr-old boys have serum cholesterol levels higher than c. have Solution: (a) Solution: Let X = (measuring) the serum cholesterol level in Let 14-yr-old boys 14-yr-old X ~ N( µ = 170 ,σ² = 900) 900) P[ 160 < X < 200] P[ = 160 − 170 200 − 170 P[ 30 <Z< 30 = P[-0.33 < Z < 1] P[-0.33 = 0.1293 + 0.3413 (from table A) 0.1293 = 0.4706 Solution:(b) Solution:(b) Let Y = (counting) the number of 14-yr-old Let boys having serum cholesterol levels between 160 and 200. between Y ~ Bin ( n=10, p= 0.4706) from (a) n=10, P[Y=7] = ( )(0.4706) (0.5294) 10 7 = 0.091 0.091 7 3 (with calculator) Solution: (c) Solution: Look up 0.49 in table B, gives z = 2.326 Look 2.326 Now X − µ , find x=c z= c - 170 170 c ∴c σ = 30 x 2.326 = 30 x 2.326 + 170 30 = 239.78 C: Normal Approximation to Binomial Distribution Distribution For X ~ Bin (n, p) the binomial probability is given by P( X = x) = ( )p q n x x n− x , x = 0,1,....., n () Now, when n is large, say, n≥30, the computation of n Now, the x is somewhat tedious. We look for approximation methods. There are two methods. The first one is the case of n is large and p is somewhat large close to 0.5, the closer the better. close the Actually, 0.1 < p < 0.9 Actually, Now, when p is close to 0.5, the distribution is somewhat symmetric, see Case (b) of Example 4. Whenever we have a symmetric distribution, we think of normal distribution. Now, we know from Example 5 that the formulas for the mean and variance of X ~ Bin (n, p) are μ= np and σ² = npqc, ~ N (np, npq ) npq respectively. X Therefore, we use to match Therefore, X ~ Bin (n, p) and use Table 1 to find the Bin required probabilities. required Xc = continuous version of X. continuous X. Normal Approximating Binomial Theorem (NABT) (NABT) For X ~ Bin (n, p) where n is large For Bin and p is not too remote from 0.5 in such a way that np ≥ 5, nq ≥ 5, and 0.1 < p < 0.9, and X can be approximated by X c ~ N (np, npq ) (Proof Omitted) (Proof Example 7: Example A fair, ordinary dice is rolled 18 times. What is the probability of getting at least 4 but NOT more than 10 “red” results? NOT Solution: On an ordinary fair dice, faces “1” and “4” are red in colour. are Therefore, on each roll the chance of Therefore, getting “red” is 2/6=1/3. getting Let X = the number of times “red” occurs when rolling the dice X ~ Bin (n =18, p =1/3) where Bin μ = np = 6 and σ² = npq = 6x2/3 = 4 Now by NABT, X is approximated by X c ~ N (np, npq) X exact Xc ~ 1 Bin(n = 18, p = ) 3 approx ~ N ( µ = 6, σ 2 = 4) Xc Now, we are required to find P(4 ≤ X ≤ 10) = P(3.5 ≤ X c ≤ 10.5) 10) That 4 ≤ X ≤ 10 has been changed into That 3.5 ≤ X c ≤ 10.5 is called the continuity correction. continuity The rule is: The ½(the last number included) +½ (tthe first number he ½( excluded) excluded 1 1 ×10 + ×11 = 10.5 2 2 which is equivalent to which for the right cut-off value for 1 1 × 4 + × 3 = 3.5 2 2 and to and for the left cut-off value. Now the Signal/Sigma ratio corresponding to 3.5 Now and 10.5 are, respectively, z1 = 3.5 − 6 = −1.25 2 and and z2 = 10.5 − 6 = 2.25 2 Hence Hence P (4 ≤ X ≤ 10) = P (3.5 ≤ X c ≤ 10.5) = P (−1.25 ≤ Z ≤ 2.25) = 0.3944 + 0.487 = 0.8822 Note: The two quantities, μandσ², characterizing a normal ², distribution, N(μ,σ²), are called parameters. distribution, ²), parameters. The two quantities, n and p, characterizing a binomial characterizing distribution, Bin (n, p), are also parameters. Bin parameters See Example 6 above. See ...
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