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HW2 assignment 1. 2. 3. A fair coin is tossed 5 times. Find a probability of obtaining (a) exactly 3 heads; (b) at most 3 heads Solution: Binomial distribution. (a) P(n=3) = C5,3 0.530.52 = Binomial[5,3]*0.55 = 0.31 (b) P(n<=3) = 1-P(4)-P(5)= 0.81 A hat contains tags numbered 1,2,3,4 and 5. A tag is drawn from the hat and replaced, then the second tag is drawn. (a) show the sample space; (b) Find the probability that the number on the second tag exceeds the number on the first. Solution: (a) S includes 25 points, {x,y}, Similar to a sample space for two dice with 5 faces each (b) There are 10 points in S where y>x. P(y>x) = 10/25 = 2/5. (c) Find the probability that the first tag has a prime number and the second tag has an even number (1 is not considered a prime number) Solution: The points are : {22},{24},{32},{34},{52},{54}. P=6/25 = 0.24. There are 4 roads connecting the cities A and B and 3 roads between the cities B and C. How many ways can round trip be made from city A to C (through B) and back? How many ways if it is desired to take different routs on the way back? 4. (a) How many license plates are possible if the first three places are occupied by letters and the last three by numbers (b) Assuming all combinations are equally likely, what is the probability the three letters and the three numbers are different? Solution (a) 263 103 (b) P26,3P10,3/ 263 103 = 0.64 5. Six students, three boys and three girls, line up in a random order for a photograph. What’s the probability that the boys and girls alternate? Solution N= 3!*3! + 3!*3!= 72; P = 72/6!= 72/720= 0.1 6. Ten children are to be grouped into two clubs (Lions and Tigers) with 5 children in each club. Each club is then to elect a president and a secretary. In how many ways can this be done? Solution C10,5P5,2P5,2=100800. 7. N people are arranged in a line. What is the probability that 2 particular people, say A and B, are not next to each other? Solution: Consider two people as a unit. There are (N-1)! Ways to permute such a group. The total number of permutations corresponding to A and B placed together is 2*(N-1)!. The probability P(A and B are together) = 2(N-1)!/N! = 2/N. The answer is P(A and B separated) = 1 – 2/N = (N-2)/N 8. How many ways can 4 couples sit at a round table if each man sits next to two women neither one of whom is his wife Placing men in every other chair (even or odd) results in 2 4! possibilities. Thus, we have 48 possible arrangements. For each of these arrangement, for every empty chair there are two possible choices of a woman per such that she is married to none of her two neighbors. Solution (1) Let’s place men first. The two possibilities are (a) every other odd chair; (b) every other even chair. The total number of arrangements is 2*4! = 48; (2) The second step is to place the women in the empty chairs. For each placement of the men, there are two possible arrangements of the women that satisfy the conditions. Consider an example: {M1 W3 M2 W4 M3 W1 M4 W2} {M1 W4 M2 W1 M3 W2 M4 W3} The answer is N=48*2 = 96. 9. How many times should you roll a die so that the probability of at least one 6 is at least 0.9? Solution P(m,n) is the probability of m six’s in n rolls. P(m>=1,n) =1-P(m=0,n); P(0,n) = (5/6)n This probability must be less or equal to 0.1 (so that the probability of the compliment event, “at least one 6”, is >= 0.9. Thus we have to find an integer n with which P(0,n) is the number closest to 0.1 from below. We can first solve the equation (5/6)n = 0.1; n = Log[0.1]/Log[5/6]= 12.6; The closest integer solution is 13. You can also do it with Mathematica by making a table of P(0,n) for {n,0,13} 10. A student has no knowledge whatsoever of the material to be tested on a true-false examination, and so the student flips the coin in order to determine the response to each question. What is the probability that the student scores at least 60% on a ten-item examination? Solution: “At least 6” means : 6 OR 7 OR 8 OR 9 OR 10. The events are disjoined and the probability equals the sum of the individual probabilities: 0.510 (Binomial[10,6]+Binomila[10,7]+… +Binomial[10,10])=0.377 11. If 70% of the bolts produced by a machine are perfect, 20 % are of average quality and 10 % are defective, determine the probability that out of 15 bolts chosen at random, (a) exactly two are average, (b) 10 are perfect, 3 are average and 2 are defective, (c) less than 3 are defective. 11. Solution (a) Exactly two are “average”. In this case the problem is reduced to Binomial distribution with m=2, p=0.2,q=0.8 P(2 average) = Binomial[2,15] (0.2)^2 * (0.8)^13= 0.23 (b) 10 are perfect, 3 are average and 2 are defective. Now we have three outcomes, and have to apply a multinomial distribution P(10,3,2) = (15!/10!3!2!) 0.7100.230.12 = 0.07 (c) Less than 3 are defective. Binomial, with p =0.1 and q = 0.9. P(ndef<3) = P(0)+P(1)+P(2) = Binomial[15,0]0.915 + Binomial[15,1]0.9140.1+ Binomial[15,2]0.9130.12 =0.82