Assign_Sol_2

Assign_Sol_2 - Assignment 2 Solutions; MATH 203 Problem...

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Assignment 2 Solutions; MATH 203 Problem 3.24 (a) The sample points are a family with 2 Boys, a family with Boy and then a Girl, a family with a Girl then a Boy, and a family with a 2 Girls. (b) If having a Boy is equally likely to having a girl and vice versa, then: Pr ( Boy - Boy ) = 1 4 ; Pr ( Boy - Girl ) = 1 4 Pr ( Girl - Boy ) = 1 4 ; Pr ( Girl - Girl ) = 1 4 . (c) We can estimate the probabilities of these intersections with their sample prob- abilities. ˆ p Boy-Boy = 1085 4208 = 0 . 258 ˆ p Boy-Girl = 1086 4208 = 0 . 258 ˆ p Girl-Boy = 1111 4208 = 0 . 264 ˆ p Girl-Girl = 926 4208 = 0 . 220 (d) There is not a large change in the probabilities from what we would expect under equal probabilities. There definitely does not seem to be a bias for people who have boys first to have boys second. There seems to be a bit of a drop for people who have girls first to have boys second. This might indicate that more that the overall probability of having boys is slightly higher, but not that it “runs in the family”. 1
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Problem 3.32 (a) Let the first letter in a gene pair represent the gene from the first parent and the second letter represent the gene from the second parent. There are four possible sample points for the gene pair for the child in the case of part (a): BB,bB,Bb,bb. Because each of these sample points has probability 1/4, we know the probability of a blue-eyed child is 1/4. If you assume independence of the parents (implied by the problems), if you let A be the event the first parent contributes a b and C be the event that the second parent contributes a b , then Pr ( bb ) = P ( A C ) = P ( A ) P ( C ) = 1 2 × 1 2 = 1 4 . (b) In this example, we have possible combinations (denoting the two genes for the second parent as b 1 and b 2 ): Bb 1 , bb 1 , Bb 2 , bb 2 . Therefore, the probability of a blue-eye child is now 1/2. Also can show using the events of part (a) with Pr ( bb ) = Pr ( A C ) = Pr ( A ) Pr ( C ) = (1 / 2)(1) = 1 / 2. (c) Because one parent has only BB pair, the probability of getting a bb pair is 0.
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This note was uploaded on 07/15/2011 for the course MATH 203 taught by Professor Dr.josecorrea during the Summer '08 term at McGill.

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Assign_Sol_2 - Assignment 2 Solutions; MATH 203 Problem...

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