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6solution - ECEN 303 Assignment 6 Problems 1 Two fair dice...

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ECEN 303: Assignment 6 Problems: 1. Two fair dice are rolled. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; In this first case, we have p X,Y ( x,y ) 2 3 4 5 6 7 8 9 10 11 12 1 1 36 0 0 0 0 0 0 0 0 0 0 2 0 2 36 1 36 0 0 0 0 0 0 0 0 3 0 0 2 36 2 36 1 36 0 0 0 0 0 0 4 0 0 0 2 36 2 36 2 36 1 36 0 0 0 0 5 0 0 0 0 2 36 2 36 2 36 2 36 1 36 0 0 6 0 0 0 0 0 2 36 2 36 2 36 2 36 2 36 1 36 . (b) X is the value on the first die and Y is the larger of the two values; In this second case, we get p X,Y ( x,y ) 1 2 3 4 5 6 1 1 36 1 36 1 36 1 36 1 36 1 36 2 0 2 36 1 36 1 36 1 36 1 36 3 0 0 3 36 1 36 1 36 1 36 4 0 0 0 4 36 1 36 1 36 5 0 0 0 0 5 36 1 36 6 0 0 0 0 0 6 36 . (c) X is the smallest and Y is the largest value obtained on the dice. Finally, in this third scenario, we obtain p X,Y ( x,y ) 1 2 3 4 5 6 1 1 36 2 36 2 36 2 36 2 36 2 36 2 0 1 36 2 36 2 36 2 36 2 36 3 0 0 1 36 2 36 2 36 2 36 4 0 0 0 1 36 2 36 2 36 5 0 0 0 0 1 36 2 36 6 0 0 0 0 0 1 36 . 2. The number of people that enter a drugstore in a given hour is a Poisson random variable with parameter λ = 10. Compute the conditional probability that at most 3 men entered the drugstore, given that 10 women entered in that hour. What assumptions have you made? Suppose that every customer that enters the store is a men with probability p or a women with probability (1 - p ), independently of other customers. Let C be the total number of 1
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customers, M be the number of men, and W be the number of women. From deriving the splitting property of a Poisson random variable, we know that p M ( m ) = (10 p ) m m ! e - 10 p m = 0 , 1 ,... and similarly p W ( w ) = (10(1 - p )) w w ! e - 10(1 - p ) w = 0 , 1 ,... Furthermore, we know that p M | C ( m | c ) = parenleftbigg c m parenrightbigg p m (1 - p ) c - m m = 0 , 1 ,...c. We can therefore write p M | W ( m | 10) = p M | C ( m | m + 10) = parenleftbigg m + 10 m parenrightbigg p m (1 - p ) 10 m = 0 , 1 ,... As such, the probability that at most 3 men entered the drugstore can be computed as Pr( M 3 | W = 10) = p M | W (0 | 10) + p M | W (1 | 10) + p M | W (2 | 10) + p M | W (3 | 10) = parenleftbigg 10 0 parenrightbigg (1 - p ) 10 + parenleftbigg 11 1 parenrightbigg p (1 - p ) 10 + parenleftbigg 12 2 parenrightbigg p 2 (1 - p ) 10 + parenleftbigg 13 3 parenrightbigg p 3 (1 - p ) 10 .
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