hw1_print

hw1_print - -( 5 6 ) 6 = . 6651. P(at least one 6 6 in 6...

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STAT416: Probability — Spring 2011 Assignments 1 Homework #1 (Solution) January 22, 2011 1 Total possible points is 45 1. Ex. 1.5 (5 pts) . Yes, because the total possible number of initials is 26 3 < 20 , 000. 2. Ex. 1.9 (5 pts) . a) Mutually exclusive (disjoint); b) Not disjoint; c) Not disjoint. 3. Ex. 1.13 (5 pts) . P(FULL) = 1 4 · 1 3 · 2 2 · 1 1 = 1 12 . 4. Ex. 1.14 (5 pts) . P(at least one bad item) = 1-P(no bad item)= 1 - 3 2 1 1 2 1 4 2 2 1 3 1 = 1 - 1 6 = 5 6 . 5. Ex. 1.15 (5 pts) . P(members of each family happen to be together) = 3! · 3! 9 · 8 · 7 3! 6 · 5 · 4 3! 3 · 2 · 1 = (3!) 4 9! . 6. Ex. 1.24 (5 pts) . P(North gets an odd number of spades) = P (1 S ) + P (3 S ) + P (5 S ) + P (7 S ) + P (9 S ) + P (11 S ) + P (13 S ) = 1 2 = 0 . 5 . ± 13 1 ² ± 39 12 ² + ± 13 3 ² ± 39 10 ² + ± 13 5 ² ± 39 8 ² + ± 13 7 ² ± 39 6 ² + ± 13 9 ² ± 39 4 ² + ± 13 11 ² ± 39 2 ² + ± 13 13 ² ± 39 0 ² ± 52 13 ² Or, there are 7 odd cases and 7 even cases (including zero), so P(North gets an odd number of spades) = 7 14 = 0 . 5. 7. Ex. 1.33 (5 pts) . P(at least one 6 in 6 rolls) = 1-P(no 6 in 6 rolls) = 1
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Unformatted text preview: -( 5 6 ) 6 = . 6651. P(at least one 6 6 in 6 rolls) = 1-P(no 6 6 in 6 rolls) = 1-( 35 36 ) 6 = 0 . 1555. (a) is more likely. 8. Ex. 1.35 (5 pts) . a) P(at least one of 3 chairs gets occupied)= 1-P(no specific chair gets occupied) = 1- 7 6 10 6 = 1-1 30 = 0 . 9667. 9. Ex. 1.43 (5 pts) . a) P(Neither north nor south gets spades) = P(east and west get all spades) = 26 13 52 13 = 0 . 00001638. b) P(N and S each has clubs and spades) = 26 13 52 13 · 13 13 39 13 = 2 . 016 × 10-15 . Purdue University hw1˙print.tex; Last Modified: January 22, 2011 (Prof. Sharabati)...
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This note was uploaded on 04/23/2011 for the course STAT 416 taught by Professor Staff during the Spring '08 term at Purdue.

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