&iacute;™•&euml;&yen;&nbsp;_&euml;&deg;_&euml;žœ&euml;&curren;&euml;&sup3;€&igrave;ˆ˜_&igrave;†”&e

# í™•ë¥ _ë°_ëžœë¤ë³€ìˆ˜_ì†”&e

This preview shows pages 1–3. Sign up to view the full content.

Problem Solutions – Chapter 5 Problem 5.1.1 Solution The repair of each laptop can be viewed as an independent trial with four possible outcomes corre- sponding to the four types of needed repairs. (a) Since the four types of repairs are mutually exclusive choices and since 4 laptops are returned for repair, the joint distribution of N 1 ,..., N 4 is the multinomial PMF P N 1 ,..., N 4 ( n 1 n 4 ) = ± 4 n 1 , n 2 , n 3 , n 4 ² p n 1 1 p n 2 2 p n 3 3 p n 4 4 (1) = ³ 4 ! n 1 ! n 2 ! n 3 ! n 4 ! ( 8 15 ) n 1 ( 4 15 ) n 2 ( 2 15 ) n 3 ( 1 15 ) n 4 n 1 +···+ n 4 = 4 ; n i 0 0 otherwise (2) (b) Let L 2 denote the event that exactly two laptops need LCD repairs. Thus P [ L 2 ]= P N 1 ( 2 ) . Since each laptop requires an LCD repair with probability p 1 = 8 / 15, the number of LCD repairs, N 1 , is a binomial ( 4 , 8 / 15 ) random variable with PMF P N 1 ( n 1 ) = ± 4 n 1 ² ( 8 / 15 ) n 1 ( 7 / 15 ) 4 n 1 (3) The probability that two laptops need LCD repairs is P N 1 ( 2 ) = ± 4 2 ² ( 8 / 15 ) 2 ( 7 / 15 ) 2 = 0 . 3717 (4) (c) A repair is type (2) with probability p 2 = 4 / 15. A repair is type (3) with probability p 3 = 2 / 15; otherwise a repair is type “other” with probability p o = 9 / 15. Define X as the number of “other” repairs needed. The joint PMF of X , N 2 , N 3 is the multinomial PMF P N 2 , N 3 , X ( n 2 , n 3 , x ) = ± 4 n 2 , n 3 , x ²± 4 15 ² n 2 ± 2 15 ² n 3 ± 9 15 ² x (5) However, Since X + 4 N 2 N 3 , we observe that P N 2 , N 3 ( n 2 , n 3 ) = P N 2 , N 3 , X ( n 2 , n 3 , 4 n 2 n 3 ) (6) = ± 4 n 2 , n 3 , 4 n 2 n 3 4 15 ² n 2 ± 2 15 ² n 3 ± 9 15 ² 4 n 2 n 3 (7) = ± 9 15 ² 4 ± 4 n 2 , n 3 , 4 n 2 n 3 4 9 ² n 2 ± 2 9 ² n 3 (8) Similarly, since each repair is a motherboard repair with probability p 2 = 4 / 15, the number of motherboard repairs has binomial PMF P N 2 ( n 2 ) n 2 = ± 4 n 2 4 15 ² n 2 ± 11 15 ² 4 n 2 (9) 205

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Finally, the probability that more laptops require motherboard repairs than keyboard repairs is P [ N 2 > N 3 ] = P N 2 , N 3 ( 1 , 0 ) + P N 2 , N 3 ( 2 , 0 ) + P N 2 , N 3 ( 2 , 1 ) + P N 2 ( 3 ) + P N 2 ( 4 ) (10) where we use the fact that if N 2 = 3or N 2 = 4, then we must have N 2 > N 3 . Inserting the various probabilities, we obtain P [ N 2 > N 3 ] = P N 2 , N 3 ( 1 , 0 ) + P N 2 , N 3 ( 2 , 0 ) + P N 2 , N 3 ( 2 , 1 ) + P N 2 ( 3 ) + P N 2 ( 4 ) (11) Plugging in the various probabilities yields P [ N 2 > N 3 ]= 8 , 656 / 16 , 875 0 . 5129. Problem 5.1.2 Solution Whether a pizza has topping i is a Bernoulli trial with success probability p i = 2 i . Given that n pizzas were sold, the number of pizzas sold with topping i has the binomial PMF P N i ( n i ) = ± ( n n i ) p n i i ( 1 p i ) n i n i = 0 , 1 ,..., n 0 otherwise (1) Since a pizza has topping i with probability p i independent of whether any other topping is on the pizza, the number N i of pizzas with topping i is independent of the number of pizzas with any other toppings. That is, N 1 N 4 are mutually independent and have joint PMF P N 1 ,..., N 4 ( n 1 n 4 ) = P N 1 ( n 1 ) P N 2 ( n 2 ) P N 3 ( n 3 ) P N 4 ( n 4 ) (2) Problem 5.1.3 Solution (a) In terms of the joint PDF, we can write joint CDF as F X 1 ,..., X n ( x 1 x n ) = Z x 1 −∞ ··· Z x n −∞ f X 1 ,..., X n ( y 1 y n ) dy 1 n (1) However, simplifying the above integral depends on the values of each x i . In particular, f X 1 ,..., X n ( y 1 y n ) = 1 if and only if 0 y i 1 for each i . Since F X 1 ,..., X n (
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 38

í™•ë¥ _ë°_ëžœë¤ë³€ìˆ˜_ì†”&e

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online