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Probability MAT Worksheet Key
1. All that remains of a bridge mix are 6 soft centers (3 mint creams, 2 orange creams and 1 lemon cream) and 4 hard centers (3 toffees and 1 nut). a. Suppose that one is chosen at random. What is the probability that i. ii. iii. iv. It is a mint cream? 3/10 It is a nut? 1/10 It is a mint cream given that it is a soft center? It is a nut given that it is a soft center? 0
3/6
For i and ii, the answer is just the number of things asked for divided by the total number of things. In iii and iv, we are told that the selection is definitely a soft center so we can ignore the hard centers. Of the soft centers, there are 6 in all of which 3 are mint creams and 0 are nuts b. Are the following events independent? i. mint and soft center? ii. nut and soft center? No No
The question basically asks whether knowing that the selection is a soft center changes the probability. The answer is yes in both cases. So the events are not independent because the answer is dependent on whether you know it is a soft center or otherwise. c. Suppose that one is chosen at random, and that a second is later chosen at random without the first being replaced. What is the probability of getting i. an orange cream followed by an orange cream? ii. a lemon cream followed by a lemon cream? iii. a hard center followed by a hard center? (2/10)(1/9)=2/90 (1/10)(0/9)=0 (4/10)(3/9)=12/90
The two probabilities (for the first and second selections) are worked out individually and then multiplied. In this case, without replacement, the two probabilities are different because on the second selection there are fewer things to pick from. d. Suppose that one is chosen at random, identified and then put back into the bag. Suppose a second is then chosen at random. What is the probability of getting i. an orange cream followed by an orange cream? ii. a lemon cream followed by a lemon cream? iii. a hard center followed by a hard center? (2/10)(2/10)=4/100 (1/10)(1/10)=1/100 (4/10)(4/10)=16/100
The two probabilities (for the first and second selections) are worked out individually and then multiplied. In this case, with replacement, the two probabilities are identical. There is no need to multiply out the probabilities on c,d; for example to write (2/10)(1/9) as 2/90. However it is important that you communicate that they are indeed multiplied.
2. All that remains of a deck of cards are the 2, 3, 5, 7, 8 and J of diamonds; the 3 and 7 of hearts, the 5, Q and K if spades and the J of clubs. (12 cards in all.) a. Suppose that one is chosen at random. What is the probability that i. ii. iii. iv. It is a diamond? 6/12 It is a 7? 2/12 It is a diamond given that it is a 7? It is an 8 given that it is red?
1/2 1/8
For i and ii, the answer is just the number of things asked for divided by the total number of things. In iii and iv, the conditional information allows us to ignore many of the cards when computing the probabilities. b. Are the following events independent? i. diamond and 7? ii. 8 and red? Yes No
The question basically asks whether having more information changes the probability. The answer is No in the first case and Yes in the second. The events in i are independent because the answer does not depend on whether you are given the extra information or not. It is the same either way! In ii the extra information does change the probability and so the events are dependent. c. Suppose that one is chosen at random, and that a second is later chosen at random without the first being replaced. What is the probability of getting i. a 3 followed by a Q? (2/12)(1/11)=2/132 ii. a red followed by a red? (8/12)(7/11)=56/132 iii. a club followed by a club? (1/12)(0/11)=0 iv. The two probabilities (for the first and second selections) are worked out individually and then multiplied. In this case, without replacement, the two probabilities are different because on the second selection there are fewer things to pick from. d. Suppose that one is chosen at random, identified and then put back into the bag. Suppose a second is then chosen at random. What is the probability of getting i. a 3 followed by a Q? ii. a red followed by a red? iii. a club followed by a club? (2/12)(1/12) = 2/144 (8/12)(8/12)=64/144 (1/12)(1/12)=1/144
The two probabilities (for the first and second selections) are worked out individually and then multiplied. In this case, with replacement, the two probabilities are identical. There is no need to multiply out the probabilities on c,d; for example to write (2/10)(1/9) as 2/90. However it is important that you communicate that they are indeed multiplied.
3. One hundred students from three residence halls were surveyed as to their favorite subject. The results obtained are given in the following table: Math English Psychology Total Allen 1 12 21 34 Wilson 0 11 15 26 Sechrist 2 10 28 40 Total 3 33 64 100
Suppose one of the surveyed students is selected at random. a) What is the probability that he/she is from Allen? 34/100 b) What is the probability his/her favorite subject is English? 33/100 In a and b, the answer is just the number of things asked for divided by the total number of things. This requires reading form the chart, but is the same idea as the in previous problems. c) What is the probability his/her favorite subject is English given that he/she is from Allen? 12/34 d) What is the probability that he/she is from Allen given that his/her favorite subject is English? 12/33 In c and d, the conditional information allows us to ignore many students when computing the probabilities. e) Are the events "from Allen" and "favorite subject is English" independent? No The question basically asks whether having more information changes the probability. The answer is Yes. The events are dependent because the answer is dependent on whether you are given the extra information or not. Note that it doesn't matter how you treat `English' and `Allen'; in c `Allen' is the extra information; in d `English' is the extra information. 4. a) A weather report describes conditions as "clear", "partly cloudy" or "overcast". What is the probability that Sometown will be described as "clear" tomorrow? Sorry, this is not a computational problem; this is conceptual. There is not enough information to answer this question. We are not told that the three possibilities are equally likely. b) The probability that Flagstaff will be "clear" tomorrow is 0.8. What are the odds in favor of clear skies? 4:1 Since the probability of `clear' is 0.8, the probability of `not clear' is 0.2. This ratio is 0.8:0.2 which after multiplying by 5 to get whole numbers gives 4:1
c) The odds against clear skies in Houghton, Michigan tomorrow are 5:1. What is the probability of clear skies in Houghton tomorrow? 1/6 The odds against are 5:1. Since 5+1=6 this ratio can be expressed as 5/6:1/6.[The point of doing this is that 5/6+1/6 =1.] As the question says `odds against' the second number is what's needed. 5. a) The probability that anyone will get this problem right is 1/20. What are the odds against someone getting the problem right? [Please forgive my lousy sense of humor.] Since the probability of `right' is 1/20, the probability of `wrong' is 19/20. This ratio is 1/20:19/20 which after multiplying by 20 to get whole numbers gives 1:19 in favor or 19:1 against. b) The odds in favor of someone smiling at my lousy sense of humor are 1:4. What is the probability of someone smiling at it? 1/5 The odds in favor are 1:4. Since 1+4=5 this ratio can be expressed as 1/5:4/5.[The point of doing this is that 1/5+4/5 =1.] As the question says `odds in favor' the first number is what's needed. 6. The peoples of the planet Ixx use an alphabet that only has 5 letters. [Seriously; I am not kidding, they really do!] a. How many 5 letter words are possible with this alphabet? (Assume that all possibilities give a word) 55=3125 b. How many 5 letter words are possible in which no letters are repeated? 5!=120 c. If a 5-letter word is chosen at random, what is the probability that it has no repeated letter? 120/3125 For a and b, we need to go through the word letter-by-letter asking how many choices each time and then multiplying everything out. In a. (just like with replacement) there are 5 chocies for each letter so 5 times 5 times...