slides23 - Lecture Stat 302 Introduction to Probability...

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Lecture Stat 302 Introduction to Probability - Slides 23 AD April 2010 AD () April 2010 1 / 17
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Exercise 1 Alf and Beth are two UBC students. They must take 4 300-level courses from a list of 12 possibilities. If they select their courses independently and at random, what is the probability that they will have exactly two courses in common? Each student has ° 12 4 ± possible choices. So the total number of choices for both students is A = ° 12 4 ± 2 = 495 2 . The number of choices such that exactly 2 courses are in common is given by B = ° 12 2 ± ° ° 10 2 ± ° ° 8 2 ± = 83160 so the probability is P = B A = 0 . 3394 . AD () April 2010 2 / 17
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Exercise 2 Suppose that 10% of the children are left-handed. (a) In a class of 20 children, what is the probability that at least two are left-handed? (b) Suppose a school has 10 classes of 20 children. If you check the classes one by one, what is the probability that the °rst left-handed child will be found in the fourth class? AD () April 2010 3 / 17
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Exercise 2 (a) Let X denote the number of left-handed children in a class of 20 children. X follow a binomial distribution of parameter n = 20 , p = 0 . 1 so P ( X ± 2 ) = 1 ² P ( X = 0 ) ² P ( X = 1 ) = 1 ² ( 1 ² p ) 20 ² ° 20 1 ± ( 1 ² p ) 19 p = 0 . 6083 (b) Let α be the probability that there is at least one left-handed child in a class of 20, then α = P ( X > 0 ) = 1 ² ( 1 ² p ) 20 = 0 . 8784 . Let Y denote the number of classes you need to check to °nd a left-handed child, Y follows a geometric distribution of parameter α so P ( Y = 4 ) = ( 1 ² α ) 3 α = 0 . 0016 . AD () April 2010 4 / 17
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Exercise 3 To estimate the number of trout in a lake, we caught 50 trout, tagged them and released them back in the lake. Later, we caught 40 trout and found out that 4 of them were tagged. From this experiment, estimate N , the number of trout in the lake. (Hint: Let
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