Probability Homework Section P. You draw a card from a deck of 5 cards. What is the probability it is a heart (There are cards of each suit in a deck)?. A committee of four is chosen at random from a group of eight women and nine men. Find the probability that the committee contains two women and two men.. Suppose a committee of boys is to be chosen from 5 boys. Al, Bill, Carl, Dan, and Elmer. Find the probability that Dan and Elmer are both on the committee. 4. A pair of fair dice are tossed. Find the probability that the sum is 7. 5. A bag contains 8 red balls, 5 white balls, and blue balls. A ball is chosen at random from the bag. a) Find the probability of choosing a red ball. b) Find the probability of choosing a ball that is not red. 6. Three cards are drawn at random from a standard deck of 5 cards. What is the probability that the cards are kings and a queen? 7. A test consists of 0 true-false questions. What is the probability of getting at least 9 correct? 8. You are dealt 5 cards from a standard deck of 5 cards. Find the probability of being dealt of a kind ( cards that are the same value and cards that are of two other different values). 9. You roll two dice. What is the probability of getting a sum less than 0? 0. A box contains six good light bulbs and four defective bulbs. a) Find the probability that a random sample of three bulbs contain one defective bulb. b) Find the probability that a random sample of three bulbs contain at least one defective bulb..in a recent year there were 5,77 people waiting for an organ transplant. The following table lists the number of patients waiting for the most common types of transplants> Organ Transplant Patients Waiting Heart,774 Kidney 5,05 Liver 7,90 Lung,40 Source: Coalition on Organ and Tissue Donation Assuming none of these people need two or more transplants, approximate the probability that a transplant patient chosen at random will need a) a kidney or a heart b) neither a kidney nor a heart
. In a hybrid corn research project, 00 seed were planted, and 70 of them germinated. find the empirical probability that any particular seed of this type will germinate.. In a certain state, 8,550 boys and 5,770 girls were born in 996. Find the empirical probability that one of those births, chosen at random would be: a) a boy b) a girl 4. Suppose you plan to take three courses next term. If you select them randomly from a list of twelve courses, five of which are science courses, what is the probability that all three courses you select will be science courses. 5. Jose has fifty hit singles from the fifties, including exactly one by Buddy Holly, two by The Drifters, three by Bobby Darin, four by The Coasters, and five by Fats Domino. If Jose randomly selects one hit from his collection of fifty, find the probability it will be by each of the following: a) Buddy Holly b) The Drifters c) Bobby Daren d) The Coasters e) Fats Domino 6. Hussein, Maria, and Sonja are three members of a class of students. Three class members will be chosen at random to deliver their reports during the next class session. a) What is the probability that Hussein, Maria, and Sonja are selected in that order. b) What is the probability that Hussein, Maria, and Sonja are selected in any order. 7. In a certain city, the air quality has been unfavorable on 7 out of 9 days. What is the probability that the air quality will be unfavorable tomorrow? 8. A couple intends to have four children, what is the probability that they will have exactly two boys? (Assume that having a boy or girl is equally likely.) 9. An often quoted example of an unlikely event is that a monkey types Shakespeare s entire play Hamlet by randomly striking keys on a typewriter. Assume that the typewriter has 48 keys (including the space bar) and that the monkey is equally likely to hit any key. a) Find the probability that such a monkey will actually correctly type just the title of the play as his first word. b) What is the probability that the monkey will type the phrase To be or not to be as his first words? 0. Many genetic traits are controlled by two genes, one dominant and one recessive. In Gregor Mendel s original experiments with peas, the genes controlling the height of the plant and T(tall) and t(short). The gene T is dominant, so a plant with the genotype (genetic makeup) TT or Tt is tall, whereas one with tt is short. By a statistical analysis of the offspring in his experiments, Mendel
concluded that offspring inherit one gene from each parent and that each possible combination of the two genes is equally likely. If each parent has the genotype Tt, then the following chart gives the possible genotypes of the offspring: Parent Parent T t T TT Tt t Tt tt Find the probability that a given offspring of these parents will be a) tall or b) short.. Make a chart of the possible genotypes of the offspring if one parent has genotype Tt and one parent has genotype tt. Find the probability that a given offspring will be a) tall or b) short.
Answers:. n(s) = 5; n(e) = ; so P(heart) =.. n(s)= C(7,4) = 80; Choose women and then choose men, n(e) = C(8,)*C(9,) = 8 6 = 008; so P( men & women) =.. 0 4. n(s) = 6; n(e) = 6; so P(sum of 7) =. 5. a) ; b). 6. n(s) = C(5,) =,00; n(e) = C(4,) C(4,) = 6 4 = 4; so P( kings & queen) =. 7. n(s) = 0 = 04; 9 correct: choose one to be wrong, C(0,) = 0; 0 correct, only one way; n(e) = 0 + = ; so P(at least 9 correct) =. 8. Choose kind for three of a kind: C(,) = ; and then choose of that kind (4,) = 4; and then choose two other kinds for the cards left: C(,) = 66; and then choose one card of each of those kinds: C(4,) C(4,) = 6. So n(e) = 4 66 6 = 54,9. P( of a kind) =. 9. n(s)= 6*6 = 6 n(e) = sum of,,4,5,6,7,8,9 = +++4+5+6+5+4 = 0; so: P(sum < 0) =. 0. a) C(6,)* C(4,) C(0,) 5*4 0 b) Need to find the probability of one defective or two defective or three defective. Add because they are separated by the work or. One defective(above): = ; Two defective: C(6,)* C(4,) C(0,) 6*6 0 0 Three defective: C(4,) C(0,) 4 0 0 So: 0 0 5 0 9 0 0 5 0 5 6 4
. a) ; b).4. 70/00 = 7/0 0.850. a) 0.59 b) 0.48 4. n(s) = C(,) ; n(e) = C(5,). So P(E) = 0/0 = /. 5. a) b) c) d) = e) 6. a) b) = 7. P(unfavorable) = 0.46 8. n(s) = 4 = 6, n(e) = C(4,) = 6. So P(exactly boys) = =. 9. a) n(s) = 48 6, n(e) = (there is only correct way to type each letter of the word). So P(E) = 8.8 X 0 - b) n(s) = 48 8, n(e) =, so P(E) = 5.47 X 0-0. a) P(tall) = b) P(short) =. Parent Parent t t T Tt Tt t tt tt a) P(tall) =. P(short) = 5