Math 21 Test

Transcription

1 Math 21 Test Name Show all your work for each problem in the space provided. Correct answers without work shown will earn minimum credit. You may use your calculator. 1. [6 points] The sample space of an experiment is {A, B, C, D} and P(A) =.1, P(B) =.3, P(C) =.4. a) Find P(D) PA ( ) + PB ( ) + PC ( ) + PD ( ) = 1 becausepsamplespace ( ) = 1... so P( D) = 1 and consequently PD ( ) = 0.2 b) Find P (A or C) PAorC ( ) = PA ( ) + PC ( ) PA ( C) wherepa ( C) = 0 = = [6 points] A person is selected at random from a group of people classified as high, moderate or low risks for lung cancer. The probability of an individual from each of these groups being selected is shown in the table below. Low Moderate High Male Female a) Find the probability that a male is selected. P(male)=P(low risk male)+p(moderate risk male) +P(high risk male) = = 0.42 b) Find the probability that a person of moderate risk is selected. P(moderate risk person)=p(moderate risk male) + P(moderate risk female) = = 0.73 c) Find the probability that a female of low or moderate risk is selected. P(female of low or moderate risk)=p(low risk female)+p(moderate risk female) = = 0.56

2 3. [6 points] Three people are randomly selected from a group of 6 men and 4 women and seated in a row. NOTE: Experiment is seating 3 of 10 people in a row can be done 10 x 9 x 8 ways. a) Find the probability the first two seated are men and the third a woman P(man, man, woman) = = 720 = 6 b) Find the probability the first seated is a woman, the second a man, and the third a woman P(woman, man, woman) = = = = [6 points] An urn contains 30 balls, of which 12 are red, 10 are yellow and 8 are green. If two balls are selected at random, what is the probability that both are the same color? NOTE: Experiment is to select 2 balls from a group of 30 can be done C(30,2) ways P(2 balls of the same color) = P(2red or 2yellow or 2 green) C(12,2) + C(10, 2) + C(8, 2) = C(30, 2) = = [6 points] A tray of electronic components contains 15 components, four of which are defective. If 4 components are selected, find the probability that Experiment: select 4 components from 15 a) all four are defective? P(4 defective) = C(4,4) C(11,0) 1 = C(15,4) 1365 b) three are defective and one is good? P(3 defective and 1 good) = C(4,3) C(11,1) 44 = C(15,4) 1365

3 6. [6 points] The probability that a coffee shop customer uses sugar is.45, the probability of using cream is.35 and the probability of using both is.20. Find the probability that a customer uses neither sugar nor cream. P( C ' and S ') = 1 P( C S) = 1 ( ) =.40 C.15 CS.20 S.25 C S [6 points] A committee is composed of 4 educators, 5 businessmen, and 6 retired people. Three people are selected at random. Find the probability that i) one member from each group is selected P(one educ and one bus and one ret'd) = C(4,1) C(5,1) C(6,1) 120 =.264 C(15,3) 455 b) all three are educators or all three are retired persons. P(3 educ or 3 ret'd) = C(4,3)+C(6,3) 24 =.053 C(15,3) 455 c) at least one educator is selected. P(at least 1 educator selected)= 1 - P(NO educators are selected) # ways to select 3 people from 5 business and 6 retired = 1 - # ways to select any 3 people C(4,0) C(11,3) = 1 - C(15,3) 165 =

4 8. [6 points] The membership of a college club consists of 11 freshmen, 16 sophomores, 27 junior and 54 seniors. There are 52 women in the club, of whom 6 are freshmen, 7 are sophomores, 13 are juniors and 26 are seniors. If one member is selected at random, find the probability that the member selected is Organize information: Freshmen Sophs Juniors Seniors TOTALS Women Men TOTALS a) A freshman or a sophomore. P(freshman or soph) = = = b) A man, given that the member is not a freshman. P(man not freshman)= n(non-freshman men) n(non-freshmen) 51 = c) A sophmore or a junior if the member selected is a woman. P(soph or junior woman)= n(soph or junior women) n(women) 20 = [6 points] One card is selected from seven cards numbered 1, 2, 3, 4, 5, 6, 7. Let E be the event "the number is odd" and let F be the event "the number is greater than 4." Use a test for independence to determine if E and F are independent. Events E and F are independent if P( E F) = P( E ) or P( F E) = P( F ) or P(E F)=P(E) P(F) Given set of cards numbered {1, 2, 3, 4, 5, 6, 7}, event E = {1, 3, 5, 7} and event F = {5, 6, 7}. Event E F = {5, 7}. So P(E) = 4/7, P(F) = 3/7 and P(E F) = 2/7. If E and F were independent, then = P(E F) = P(E) P(F) but 2/7 (4/7)(3/7) so these events are NOT independent using Test #3. Note: You can also use the first or the second test: P(E F) = 2/3 but this is not equal to P(E) = 4/7 so the events are NOT independent. Similarly, P(F E) = 2/4 which is not equal to P(F) = 3/7.

5 10. [6 points] The probability Dan oversleeps and misses his 8 o'clock class is 0.7 and the probability Tom independently oversleeps and misses his 8 o'clock class is 0.1. Find the probability that a) both oversleep and miss their 8 o'clock classes P(both oversleep) = P(Dan oversleeps AND Tom oversleeps) = P(Dan oversleeps) P(Tom oversleeps) = (0.7)(0.1) = 0.07 b) exactly one of these two students oversleeps and misses his 8 o'clock class. This occurs if Dan oversleeps and misses class but Tom doesn t or Dan doesn t oversleep and Tom does. Let D = event Dan oversleeps and T = event Tom oversleeps. We know P(D)=0.7 and P(T)=0.1 so P(D ) = 0.3 and P(T )=0.9. Then P(exactly one oversleeps) = P(D and T' or D' and T) = P(D) P(T' ) +P(D' ) P(T) = (0.7)(0.9) +(0.3)(0.1) = = [6 points] In an entering freshman class at a small private university, 18% of the entering freshmen attended private schools while 82% attended public schools. 30% of those who attended private schools and 15% of those who attended public schools played in their high school bands. If a student is chosen at random from this class, find the probability that the student attended public school if s/he played in their high school band. Private school.18 Band.30 Band.15 P(public sch P( public school band) = P(band) (.82)(.15) = (.82)(.15) + (.18)(.30) band) Public school.82

6 12. [6 points] An insurance salesperson makes a sale 30% of the time she calls on a prospect. Suppose the salesperson calls on 7 prospects. Find the probability she will make a sale a) in three of the seven sales calls. Use Bernoulli Formula for Binomial Probability where number of trials n = 7 calls, probability of success is p = probability of sale = 0.3, probability of failure is q = probability of no sale = 0.7. Here we are looking for the probability of x = 3 successes (sales). Px 3 4 ( = 3) = C(7, 3) (0.3) (0.7) b) in at least one of the seven sales calls. In this problem we are looking for the probability of 1 or 2 or 3 or 4 or 5 or 6 or 7 sales in 7 sales calls. We can calculate individual binomial probabilities and add but it s quicker and easier to notice that P(at least 1 success) = 1 - P(0 successes - which means 7 failures) = 1 C(7, 0) (0.3) (0.7) =