VIEWS: 7 PAGES: 14 POSTED ON: 7/25/2011
DISTRIBUTIONS 1. Toss a coin 3 times. Find the probability of getting 2 heads. 2. Draw 6 cards from a deck of 52 cards without replacement. What is the probability of getting two hearts? 3. A printer in a computer lab works 60% of the time. Find probability that the printer is working 2 out of the 10 visits. 4. Suppose you invest a fixed sum of money in each of five business ventures. Each has a 70% chance of success. Find the probability that exactly 2 of the business ventures will be successful. 5. A large population of accounts is known to have 10% of accounts in error. Calculate the probability that when accounts are chosen at random from this population, the first account chosen will be in error. 6. A rail company knows that on average they will have 3 people ill each day. In order to make plans for replacements they have asked you to calculate the following probabilities. In a single day (i) No employees ill (ii) Two or less employees ill (iii) Four or more employees ill 7. Suppose the probability that a married couple will have child with blonde hair is 0.25. If there are four children in the family, what is the probability that exactly half of them have blonde hair. 8. On the bus route that a student takes to college, there are 5 traffic lights between the stop the student boards the bus and the stop the student embarks. Assume that the probability that a light will be red is 0.25. Assuming independence, what is the probability that at least three traffic lights will show a red light before the student gets off the bus? 9. A secretary receives application forms for a current vacancy and has to check them for completeness before further processing is carried out on them. Suppose in a batch of 20, 7 are incomplete and the others are complete. If the secretary randomly selects 5 of these forms for checking, what is the probability that: i) none of the incomplete forms are selected? ii) at least 3 of these selections are incomplete? 10. Jack applied for 8 jobs as an accountant last week. The probability of being called for an interview is 0.25. Assuming independence, (i) what is the probability that Jack will be called for at most 2 interviews. (ii) on average, of the 8 jobs Jack applied for, how many interviews should Jack expect to be called for? 11. On average 2 people are killed on a particularly dangerous stretch of road each year. a) What is the probability that in a given year that (i) No one is killed (ii) 2 or less are killed (iii)Four or more are killed b) What is the probability that over a 3 year period less than 3 people are killed 12. A company tenders for a new contract each week. In the past they have won 40% of the contracts they have tendered for. In a 6-week period what is the probability that they will (i) Win all six contracts (ii) Win at least 2 contracts (iii) Win less than 3 contracts 13. The number of blemishes that occur on a roll of carpet during manufacture is Poisson distributed, with mean 0.4 blemishes per roll. What percentage of rolls are classified (i) Perfect (having no blemishes) (ii) Seconds (having one or two blemishes) (iii) Scrap (all the rest) 1. Bin 2. Hypergeometric 3. Bin 4. Bin 5. Geometric 6. Poisson 7. Bin 8. Bin 9. Hypergeometric 10. Bin 11. Poisson 12. Bin 13. Poisson DISTRIBUTIONS Binomial 1. Toss a coin 3 times. Find the probability of getting 2 heads. 2. Suppose you invest a fixed sum of money in each of five business ventures. Each has a 70% chance of success. Find the probability that exactly 2 of the business ventures will be successful. 3. A printer in a computer lab works 60% of the time. Find probability that the printer is working 2 out of the 10 visits. 4. A stock will show an increase or a decrease in its closing price on a daily basis. With this particular stock, the probability of an increase is 0.65. What is the probability that there will be at least 1 increase in the stock price in the next ten days. 5. A company tenders for a new contract each week. In the past they have won 40% of the contracts they have tendered for. In a 6-week period what is the probability that they will (iv) Win all six contracts (v) Win at least 2 contracts (vi) Win less than 3 contracts 6. Suppose the probability that a married couple will have child with blonde hair is 0.25. If there are four children in the family, what is the probability that exactly half of them have blonde hair. 7. On the bus route that a student takes to college, there are 5 traffic lights between the stop the student boards the bus and the stop the student embarks. Assume that the probability that a light will be red is 0.25. Assuming independence, what is the probability that at least three traffic lights will show a red light before the student gets off the bus? 8. Jack applied for 8 jobs as an accountant last week. The probability of being called for an interview is 0.25. Assuming independence, (iii) what is the probability that Jack will be called for at most 2 interviews. (iv) on average, of the 8 jobs Jack applied for, how many interviews should Jack expect to be called for? Hypergeometric 8. Draw 6 cards from a deck of 52 cards without replacement. What is the probability of getting two hearts? 9. A secretary receives application forms for a current vacancy and has to check them for completeness before further processing is carried out on them. Suppose in a batch of 20, 7 are incomplete and the others are complete. If the secretary randomly selects 5 of these forms for checking, what is the probability that: ii) none of the incomplete forms are selected? ii) at least 3 of these selections are incomplete? Geometric 10. A large population of accounts is known to have 10% of accounts in error. Calculate the probability that when accounts are chosen at random from this population, the first account chosen will be in error. 11. Auditions for a new musical will last on average 20 minutes. The director in charge of the audition knows from past experience that the probability that a person is suitable for the part is 0.8. Due to time constraints, the first person auditioned, that the director deems suitable, is selected. If the talent for any one person is independent of another, what is the probability that the director will have found the person for the part in exactly one hour after starting the auditions? Poisson 12. A rail company knows that on average they will have 3 people ill each day. In order to make plans for replacements they have asked you to calculate the following probabilities. In a single day (iv) No employees ill (v) Two or less employees ill (vi) Four or more employees ill 13. On average 2 people are killed on a particularly dangerous stretch of road each year. a) What is the probability that in a given year that (iv) No one is killed (v) 2 or less are killed (vi) Four or more are killed b) What is the probability that over a 3 year period less than 3 people are killed 14. The number of blemishes that occur on a roll of carpet during manufacture is Poisson distributed, with mean 0.4 blemishes per roll. What percentage of rolls are classified (iv) Perfect (having no blemishes) (v) Seconds (having one or two blemishes) (vi) Scrap (all the rest) 2006 Exam 2005 Repeat Exam Q4 2004 Exam Question 3 2004 Exam Question 4 2004 Repeat Exam Question 4 Solution to question 5 M = 13 number of hearts L = 39 number of non-hearts N = 52 total 13 39 2 4 P (2 hearts) =0.315 52 6 BINOMIAL DISTRIBUTION Q3 If the probability that any one person requires medical attention in one particular year is 0.25, what is the probability that: (i) four people live through the year without medical attention? (ii) just one of the four needs medical attention? (iii) at least two of the four require medical attention Q4 A company tenders for a new contract each week. In the past they have won 40% of the contracts they have tendered for. In a 6-week period what is the probability that they will (vii) Win all six contracts (viii) Win at least 2 contracts (ix) Win less than 3 contracts Q5 Suppose the probability that a married couple will have child with blonde hair is 0.25. If there are four children in the family, what is the probability that exactly half of them have blonde hair. POISSON DISTRIBUTION Q6 A rail company knows that on average they will have 3 people ill each day. In order to make plans for replacements they have asked you to calculate the following probabilities. In a single day (vii) No employees ill (viii) Two or less employees ill (ix) Four or more employees ill Q7 On average 2 people are killed on a particularly dangerous stretch of road each year. a) What is the probability that in a given year that (vii) No one is killed (viii) 2 or less are killed (ix) Four or more are killed b) What is the probability that over a 3 year period (i) Less than 3 people are killed Q8 The number of blemishes that occur on a roll of carpet during manufacture is Poisson distributed, with mean 0.4 blemishes per roll. What percentage of rolls are classified (vii) Perfect (having no blemishes) (viii) Seconds (having one or two blemishes) (ix) Scrap (all the rest) BINOMIAL DISTRIBUTION Q3 If the probability that any one person requires medical attention in one particular year is 0.25, what is the probability that: (i) four people live through the year without medical attention? (iv) just one of the four needs medical attention? (v) at least two of the four require medical attention We define a person requiring medical attention as a “success”, and the probability of a success equals 0.25. Using the Binomial Distribution n! X n- X P(X) = p (1 - p ) X! (n - X)! with n=4 people p=0.25 1-p =0.75 (i) No-one needs medical attention P(X=0) 4! 4! (0.25) ( 0.75 ) (1)(.32) .32 0 4- 0 P( 0 ) = 0! ( 4 - 0 )! 1.(4!) (ii) P(X =1) 4! 4.3.2.1 (0.25) ( 0.75 ) (0.25)(.42) .42 1 4- 1 P( 1 ) = 1! ( 4 - 1 )! 1.(3.2.1) (iii) At least two need medical attention P(X2) P(X2) =P(2) + P(3) + P(4) = 1 –( P(0) + P(1)) = 1 –(.32+.42) = .26 Q4 Winning a contract can be defined a success. The probability of winning a contract (p) = .4. There are six contracts (n=6). Using Binomial n! X n- X P(X) = p (1 - p ) X! (n - X)! 6! 0 6 0 P( 0 ) = .4 ( .6 ) = .047 0! ( 6 - 0 )! 6! 1 6 1 P( 1 ) = .4 ( .6 ) = .187 1! ( 6 - 1 )! 6! 2 6 2 P( 2 ) = .4 ( .6 ) = .311 2! ( 6 - 2 )! 6! 3 6 3 P( 3 ) = .4 ( .6 ) = .276 3! ( 6 - 3 )! 6! 4 6 4 P( 4 ) = .4 ( .6 ) = .138 4! ( 6 - 4 )! 6! 5 6 5 P( 5 ) = .4 ( .6 ) = .037 5! ( 6 - 5 )! 6! 6 6 6 P( 6 ) = .4 ( .6 ) = .004 6! ( 6 - 6 )! (i) Win all six contracts P(X= 6) P(X=6) = .004 (i) Win at least 2 contracts P(X2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) =(.311 + .276 + .138 + .037 + .004) = .766 OR P(X2) = 1 – P(X< 2) 1 - P(0) + P(1) = 1 – (.047 + .187) = .766 (iii) Win less than 3 contracts P(X< 3) = P(X=0) + P(X=1) + P(X=2) = .05 +.19 + .31 = .55 Q5 Probability of blond hair = .25. There are 4 kids (n=4). What is the probability that half of them (X=2) have blond hair. n! X n- X P(X) = p (1 - p ) X! (n - X)! 4! .25 ( .75 ) 6(.0625)(.625) .21 2 2 P( 2 ) = 2! ( 2 )! POISSON DISTRIBUTION Q6 (i) No employees ill, X = 0, =3 e 3 30 (.05)(1) P(0) = .05 0! 1 ii) Two or less ill P(X2) = P(0) + P(1) + P(2) e 3 (3)1 (.05).3 P(1) = .15 1! 1 e 3 (3) 2 (.05)(9) P(2) .22 2! 2 P(X2) = P(0) + P(1) + P(2)= .05 +.15 + .22 =.42 (iii) Probability of four or more employees ill P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3)) e 3 (3) 3 (.05)(27) P(3) =.22 3! 6 P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3)) = 1 – (.05 +.15 +.22 +.22) = 1 - .64 =.36 Q7 a) On average 2 people are killed (=2) (i) No-one killed, X = 0, =2 e 2 2 0 (.135)(1) P(0) = .135 0! 1 ii) Two or less killed P(X2) = P(0) + P(1) + P(2) e 2 (2)1 (.135).2 P(1) = .27 1! 1 e 2 (2) 2 (.135)(4) P(2) .27. 2! 2 P(X2) = P(0) + P(1) + P(2) = .135 +.27 + .27 =.675 (iii) Probability of four or more killed P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3)) e 2 (2) 3 (.135)(8) P(3) =.18 3! 6 P(X4) = 1 – P(X3) = 1 – (P(0) + P(1) + P(2) + P(3)) = 1 – (.135 +.27 +.27 +.18) = 1 - .855 =.145 b) In a 1-year period 2 people are killed on average this implies in a 3 year period 6 people are killed on average. (=6) (i) Prob (X < 3) e 6 (6) 0 (.002)(1) P(0) .002 0! 1 e 6 (6)1 (.002)(6) P(1) = .012 1! 1 e 6 (6) 2 (.002)(36) P(2) .036 2! 2 Prob (X < 3) = P(0) + P(1) + P(2) = .002 +.012 + .036 = .05 Q8 Mean number of blemishes = 0.4 (=0.4) (i) Perfect (having no blemishes) P(X = 0) e 0.4 (0.4) 0 (.6703)(1) P(0) .6703 (67.03% will be perfect) 0! 1 (ii) Seconds (having one or two blemishes) P(X = 1 or X = 2) e 0.4 (0.4)1 (.6703)(0.4) P(1) . 2561 1! 1 e 0.4 (0.4) 2 (.6703)(.16) P(2) .0536 2! 2 P(X = 1 or X = 2) = P(1) + P(2) = .2561 + . 0536 = .3097 (30.97% will be seconds) (iii) Scrap (all the rest) P(X > 2) P(X > 2) = 1 – (P(0) + P(1) + P(2)) = 1 – (.6703 + .2561 + .0536) = 1 - .98 = .02 (2% of all carpets will be scrap)