416-7sol

# 416-7sol - Denker Spring 2010 416 Stochastic Modeling...

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Unformatted text preview: Denker Spring 2010 416 Stochastic Modeling - Assignment 7 Solution: Problem 1: (Problem 55, p. 273) Consider a population of individuals each of whom posesses two genes which can be either type A or type a. Call type A dominat and type a recessive. Call an individual dominant if it has a dominant type, otherwise recessive. Let the population have stabilized with probabilities p , q and r . Compute the probability that the offspring is recessive in case (i) the parents are dominant and (ii) one of the parents is dominant the other recessive. Solution: We calculate the conditional probability S that the offspring is of types aa provided the parents are both dominant. The intersection of both events is the set of offsprings of types aa and two parents each of them of types Aa. The probability of this event is ( 1 2 r ) 2 . The probability that both parents are dominat is (1 − q ) 2 . It follows that S = r 2 4(1 − q ) 2 . Likewise, the conditional probability S ′ that the offspring is of types aa while one parent is dominant and the other recessive, is S ′ = 2 q 1 2 r 2 q (1 − q ) = r 2(1 − q ) ....
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416-7sol - Denker Spring 2010 416 Stochastic Modeling...

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