Test problems’ discussion

(1.1) A coin loaded to come up heads 2/3 of the time, is thrown until a head appears. What is the probability that the even number of tosses is necessary? Hints: (a) What kind of distribution should be used? Answer P = qp + q^3 p + q^5 p +… = q p ( 1 + q^2 + q^4 +…) = qp/(1-q^2) = 1/4. (1.2) A committee of 5 is chosen from a group of 8 men and 4 women. What is the probability the group contains a majority of men? Answer: A = either 3 or 4 women. P(A) = (Binomial[8,5] +Binomial[8,4]*Binomial[4,1])+ Binomial[8.3] Binomial[4,2])/Binomial[12,5] = 0.85 (1.3) A box contains tags numbered 1,2,…, n. Two tags are chosen without replacement. What is the probability they are consecutive integers? Hint: How many pairs satisfy this condition Answer: 2 (n-1)/Binomial[n,2] = 2 (n-1) (n-2)!/n!= 2/n. The total probability is 2/n.

(1.4) If 20% chips produced by a machine are defective, find the probability that out of 5 chips chosen at random (a) 1 (b) at least 2 chips will be defective. Solution: (a) 5*0.2*0.84 =0.41 (b) 1 – *0.2*0.84 = – 0.41 = In how many ways can 8 people be seated at a round table if (a) they can seat anywhere (b) three of them, Bob, Michael and Jeff, should not sit together (in other words all the arrangements where B, M and J sit in any order in three consecutive chairs must be excluded). Note: Only relative positions are important. Solution Fix the position of one person and rearrange all other people relative to him. There will be (8-1)! = 7! = 5040 possible permutations. Fix the position of one person not belonging to the group of three. Find first the arrangements keeping B, M and J together. Consider them a unit. It leaves us with 5 objects and 5! = 120 permutations. There is also 3! permutations within the group of 3. Thus, the total number of permutations with M B J sitting together is 120*6 = 720. The answer is: 5040 (the total number of permutations) – 720 (permutations keeping B, M and J together) = 4320.

2.1 A boy has six coins, each of different denomination. How many different sums of money can he form? Answer: 2^6 -1 = 63 or Binomial[6,1] + Binomial[6,2] +,,+Binomial[6,6] = 63. (2.2) The probability that an entering college student will graduate is 0.6. Find the probability that out of 5 students (a) none (b) at least two (c) at least one will graduate Solution: p(none) =p(0) = 0.45 = (b) 1 – p(0) – p(1) = 1 – p(0) – 5*0.6*(0.4)4=0.91 ( c ) p(n>0) = (2.3) A box contains tags numbered 1,2,…, 20. Two tags are chosen without replacement. What is the probability their sum is even? Hint: How many even/odd numbers among the tags? How can an even number be composed of two even or odd numbers? Answer: EvenSum=(Even1,Even2) OR (Odd1,Odd2); P= 1/2*(9/19) + ½*9/19 = 9/19.

(2.4) A committee of 6 is chosen from a group of 9 men and 5 women. What is the probability the group contains either four men or four women. P(A) = (Binomial[9,4]*Binomial[5,2]+Binomial[9,2]*Binomial[5,4])/Binomial[14,6] ~ (2.5) A coin loaded to come up heads ¼ of the time, is thrown until a head appears. What is the probability that the odd number of tosses is necessary? Hints: (a) What kind of distribution should be used? Answer P = p + q^2 p + q^4 p … = p /( 1-q^2) = ¼ *16/7= 4/7.

(3.1) A coin loaded to come up heads 1/3 of the time, is thrown until a head appears. What is the probability that at least 3 tosses is necessary? Hints: (a) What kind of distribution should be used? Solution P(n>2) = q^2 p + q^3 p + q^4 p… = p q^2/( 1-q) = (1/3) *(4/9)*3 =4/9. And here is the alternative and simpler solution : P (n>2) = 1 – p(1)-p(2) = 1-1/3 – 2/9 = 4/9 (3.2) How many straight lines are determined by 8 points, no 3 of which are collinear. Answer: Binomial[8,2]= 28

(3.3) Three integers are selected at random from the set {1,2,.. 10}. What is the probability that the largest of them is 5? Hint: To fit the condition, there are two groups of objects to chose from. What are they? What is the distribution to use? Solution: The groups are {5} (1 object) and {1,2,3,4} (4 objects). To find the probability that one object is chosen from the first group and one from the second, we use the hypergeometric distribution (see Lecture 3): P = Binomial[1,1] Binomial[4,2]/Binomial[10,3] = 1/20. (3.4) What is the probability of getting a total of 9 (a) twice (b) at least twice in 6 tosses of a pair of dice. Answer (a) p(2) = Binomial[6,2](1/9)2 (8/9)4 = (b) p(n>1) = 1 – p(0) – p(1) = 1 – (8/9)^6 – 6 (8/9)5(1/9) =

(3.5) How many 4 digit numbers can be formed with the 10 digits 0,1,2,3…,9 if (a) repetitions are allowed (b) the repetitions are not allowed (c) the last digit must be 0 and the repetitions are not allowed. Answer (a) The first digit can be chosen in 9 ways since 0 is not allowed. Other 3 can be any of 10. Answer: (b) 9*9*8*7 = 4536 (c) 9*8*7 = 504.