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a. Canadian background = Number of people with Canadian background/total number of people = 513/ = 0. b. African background = Number of people with African background/total number of people = 72/ = 0. c. Other background = Number of people with ‘other’ background/total number of people = (1000-513-148-72-56)/ = 211/ = 0.
a. There are 3 regions; A, B and B, the favourable outcome is spinning A out of the 3 regions. Probability = number of favourable outcomes/ total number of possible outcomes = 1/
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Therefore, the probability of spinning A is 1/3 or 0.3 or 33.33% b. c. Looking at the tree diagram there are 2 favourable outcomes of
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a. Binomial i. This is used when there is a fixed number of trials and the trial has two outcomes. ii. The formula is based on the fixed probability of success since the events are independent and the distributions tells us the probabilities of different numbers of success. The formula is created using the total number of trials, probability of success and failure as well as the number of successes and failures. iii. 20 students taste the new ice coffee at Tim Horton’s, the probability of them liking it is 75%. You can use the formula to determine the probability of each student liking the drink from 0 liking it to 25 liking it. b. Geometric i. This is used when there is not a fixed number of trials but still each trial has 2 outcomes. ii. There is a fixed probability of success. The formula is based on the probability of failure, the probability of success and the
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number of failures in order to determine the probability of success on the k th trial. iii. Kiss 92.5 is running a contest, each time you call in you are entered into a draw to win a car. The probability of winning is 0.05, if you call 30 times what is the probability of winning? c. Hypergeometric i. This is used when we are picking a fixed number from a finite pool, the events are dependent, it is for when you want to distinguish between items of a type we are interested in and those we are not. ii. The formula helps us get the probability of our sample containing certain numbers of the type we are interested in. the formula is based on how many of the type we want, how many we pick of the type we want, how many of the type we don’t want, how many we pick of the type we don’t want, the total in the pool and how many we pick in total. iii.My fruit salad has 6 blueberries and 4 raspberries; I pick 3 fruits, what is the probability of me picking blueberries?
a. RED BLUE DIE
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The probability of rolling a sum greater than 10 is 3/36, or 0.0863 or 8.3 % ii. We can count on the table where the rolling a red die is one greater than on the blue die (2,1) (3,2) (4,3) (5,4) (6,5). There are 5 favourable outcomes = 5/ =0. =13.8 % The probability of rolling a number on the red die that is one larger than on the blue die is 536, or 0.138 or 13.8% iii. If we look at the table, all sum of the 2 numbers add to less than 11 except for 3 (5,6) (6,5) and (6,6). There are 33 favourable outcomes = 33/ =0. =91.6% The probability of rolling 2 numbers that the sum is less
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than 11 is 33/36 or 0.916 or 91.6%
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- Since there are 8 copies of each style of book, if Jack chooses a novel there are 23 books left for him to choose from. From the 23 books 16 of them are not novels, the probability of picking a different style book is 16/23 or 0. or 70%
- N = 4 trials Success probability (p) = 0.8 (4/5) P(X-x) = n!/(x!(n-x)!)p^x(1-p)^(n-x) X= 0,1,2,3,..n np = 3. n * p (1-p) = 0. P (X=0) = 0. P (X=1) = 0. P (X=2) = 0. P (X=3) = 0. P (X=4) = 0.
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The probability she will score over 90 on the 3 of tests is 0.4096 or 40%
p = probability the laptop is faulty q = probability the laptop is NOT faulty p = 0. q = 1-0. = 0. p (x) = c(n,x) * p^x * q ^ (n-x) c (n,x) = n!/ (x! * (n- x)!) n = 30 p = 0. q = 0. P (2) = c (30,2) * (0.05^2) * (0.95^28) = 435 * 0.0025 *0. = 0. = 25.86% a. The probability that 2 laptops are faulty is 25.86% b. The probability that more than 2 are faulty is the probability of 2 minus 1.
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b. P (0) = (7/13) (7/13) (7/13) (7/13) = 2401/ = 0. Probability of no cat’s chosen is 0.0840 Probability of atleast 1 cat chosen is = 1- 0. = 0. Probability of atleast 1 cat chosen is 0.9160 or 91.60% 10. a = 0. a) he will for the first time in his fifth shot = he fails to shoot on shots 1- = (1-0.18) ^ = 0. b) takes less than 3shots = a + (1-a) = 0.18 + 0.82 * 0. = 0. = 32.76%
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c) how many shots before scoring = a (1-a) ^n E(X) = from n=1 to infinite of (n a (1- a)^n) E (X) = 1/a = 1/0. =5.
P (correct answer) = 1/5 P (wrong answer) = 4/ a. P (x=4) = (1/5) ^ 4 = 0. =0.16%
