This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SIEO 3658 Assignment #5 Solutions Probability October 30, 2011 Prof. Mariana OlveraCravioto Page 1 of 4 Assignment #5 Solutions 1. profit = 100 X + 200 Y . The mean of this given by E [profit] = 100 E [ X ] + 200 E [ Y ]. Since for X , each outcomes has the same probability, the mean of X is just the mean of the outcomes, i.e. EX = 2 + 4 2 = 1. For Y , we have E [ Y ] = 3 × 1 21 + 2 × 2 21 + 1 × 3 21 + 0 × 3 21 + ... + 3 × 3 21 + 4 × 2 21 + 1 × 1 21 = 1 . So the expected profit is 100 + 200 = 300. 2. (a) i. P ( I = i,J = j ) = 1 n X i =1 m i ii. P ( I = i ) = m i n X i =1 m i iii. P ( I = i ) = n X i =1 1 ( j ≤ m i ) n X i =1 m i (b) m i X j =1 [ p ij a + (1 p ij ) b ] 3. 1 ( 43 6 ) + ( 6 1 )( 43 5 ) + ( 6 2 )( 43 4 ) ( 49 6 ) 4. (a) First observe that X = x is equivalent to having all three X i ’s at least x , but not all of them can be striclty greater than x . That is, for x = 101 ,..., 110, P ( X = x ) = P ( X 1 ≥ x ) P ( X 2 ≥ x ) P ( X 3 ≥ x ) P ( X 1 > x...
View
Full
Document
This note was uploaded on 12/11/2011 for the course IEOR 3658 taught by Professor Olvera during the Fall '08 term at Columbia.
 Fall '08
 Olvera

Click to edit the document details