TEST1_F22_STAT230_SOLUTIONS (B)

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School

University of Waterloo *

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Course

230

Subject

Statistics

Date

Jan 9, 2024

Type

pdf

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{ Special instructions • Your final numerical answers should be given to 3 decimal places (e.g. 0.329). However, between steps/parts you should carry more decimal places to avoid rounding errors. • If working is not specifically requested, a correct answer will receive full marks. However, an incorrect answer could still receive credit if working is provided. An incorrect answer with no working will receive zero marks. • The exam consists of 4 questions for a total of 35 marks. The number of marks available per question is indicated in [square brackets]. • Answer the questions in the spaces provided. You may use the last page of the test for any additional work you would like to be graded. If you use this space, make it as clear as possible which question(s) your work relates to.

Question 1 [8 marks] Annie’s children are playing with blocks by arranging them in a straight line. One of the children is given 4 red blocks, 5 black blocks, and 3 yellow blocks. Given the child is a toddler, we assume the blocks are placed randomly in the line. (a) [2 marks] How many block patterns can be created, if we consider blocks of the same colour are indistinguishable? We have 12 blocks total to arrange with 4 indistinguishable red blocks, 5 indistinguishable black blocks, and 3 indistinguishable yellow blocks. Thus, the total number of arrangements is 12! 4!5!3! = 27 , 720. (b) [2 marks] Find the probability that there is a yellow block at each end of the line. We can fix a yellow block at each end as Y x x x x x x x x x x Y. The remaining 10 blocks can be arranged in the middle ‘x’ places in 10! 4!5!1! = 1260 ways. Thus, the probability is 1260 27720 = 0 . 0455. (c) [2 marks] Find the probability that all 4 of the red blocks are together. Consider the 4 red blocks as one object and then arrange the 9 items of RRRR B B B B B Y Y Y. This can be done in 9! 1!5!3! = 504 ways. So the probability the 4 R’s are together is 504 27720 = 0 . 0182. (d) [2 marks] Find the probability that there is a yellow block at each end and all of the red blocks are together. We cannot apply the multiplication rule and multiply the number of outcomes in parts (b) and (c) to- gether because the number of ways to arrange the group of red blocks is dependent on the arrangement of the yellow blocks. Instead, fix a yellow block at each end as before and then arrange the remaining 7 blocks of RRRR B B B B B Y in 7! 1!5!1! = 42 ways. Thus, the probability is 42 27720 = 0 . 00152.

Question 2 [8 marks] Ten students (including Abby and Becky) have applied for the same junior data analyst positions at Amazon. Amazon hires at random 4, 5 or 6 of the students. (a) [2 marks] How many possible ways are there to select the students to hire? Since order does not matter, the number of possible ways is ( 10 4 ) + ( 10 5 ) + ( 10 6 ) = 210+252+210 = 672. (b) [2 marks] Calculate the probability that Abby is hired. If Abby is hired, we need to find the remaining students from the remaining 9 people. So there are ( 9 3 ) + ( 9 4 ) + ( 9 5 ) = 84 + 126 + 126 = 336 possible ways for other students to be hired. So the probability is 336 / 672 = 0 . 5 . (c) [2 marks] Calculate the probability that either Abby or Becky (but not both) is hired. We found P ( A ) = 0 . 5 in b). By symmetry, P ( B ) = 0 . 5. The probability of both Abby and Becky getting hired is P ( A ∩ B ) = ( 8 2 ) + ( 8 3 ) + ( 8 4 ) 672 = 154 / 672 = 0 . 229 . We want P ( A ∪ B ) − P ( A ∩ B ) = P ( A )+ P ( B ) − P ( A ∩ B ) − P ( A ∩ B ) = 0 . 5+0 . 5 − 0 . 229 − 0 . 229 = 0 . 542 . (d) [2 marks] Calculate the probability that neither Abby nor Becky is hired. We want P ( A c ∩ B c ) = P ( A ∪ B ) c = 1 − P ( A ∪ B ) = 1 − P ( A ) − P ( B )+ P ( A ∩ B ) = 1 − 0 . 5 − 0 . 5+0 . 229 = 0 . 229 . Alternate solution: exclude Abby and Becky and pick from the remaining 8 people: ( 8 4 ) + ( 8 5 ) + ( 8 6 ) 672 = 154 / 672 = 0 . 229 .

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