Solutions to Problem Set 5

# Solutions to Problem Set 5 - b. P ( one and only one 3 ∩...

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Solutions to Problem Set 5 February 28, 2012 Problem 1 a. There are 2 4 possible outcomes. (writing down all of them are time consum- ing and are omitted here.) b. easy c. P ( A ) = 2 ( 1 2 ) 4 = 0 . 125 P ( B ) = ± 4 1 ² ( 1 2 ) 4 = 0 . 25 P ( C ) = 1 - P ( C c ) = 1 - ± 4 1 ² ( 1 2 ) 4 - ± 4 0 ² ( 1 2 ) 4 = 0 . 6875 Problem 2 It is easier to answer the following questions by ﬁrst writing down the con- tingency table: males females row sum group 1 0 . 07 0 . 4 0 . 47 group 2 0 . 33 0 . 2 0 . 53 column sum 0 . 4 0 . 6 1

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a. 0.4 b. 0.6 c. 0.47 d. P ( male group 1 ) = P ( male ) + P ( group 1 ) - P ( male group 1 ) =0.4+0.47- 0.07=0.8 e. 0.4 f. 7/40 Problem 3 The i-th dice is represented by X i . a. P ( X 1 + X 2 + X 3 = 7 X 1 X 2 X 3 ) = P ( X 1 + X 2 + X 3 = 7 X 1 X 2 X 3 ) P ( X 1 X 2 X 3 ) P ( X 1 X 2 X 3 ) = ± 6 1 ²± 5 1 ²± 4 1 ² 6 3 2
The events satisfying X 1 + X 2 + X 3 = 7 X 1 X 2 X 3 are the permutations of { 1 , 2 , 4 } . Therefore, P ( X 1 + X 2 + X 3 = 7 X 1 X 2 X 3 ) = 3! 6 3 . It follows that the probability of the event of interest is given by 1/20.
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Unformatted text preview: b. P ( one and only one 3 ∩ X 1 ≠ X 2 ≠ X 3 ∣ X 1 ≠ X 2 ≠ X 3 ) = ± 3 1 ²± 5 1 ²± 4 1 ² 6 3 ± 6 1 ²± 5 1 ²± 4 1 ² 6 3 = . 5. Problem 4 a. 4 15 5 15 6 15 = . 0356 . b. 4 15 5 14 6 13 = . 04395 . Problem 5 a. P ( only one ) = P ( only Paul ) + P ( only Charles ) + P ( only David ) = 1 6 3 4 2 3 + 5 6 1 4 2 3 + 5 6 3 4 1 3 = . 43 b. P ( only Paul ) P ( only one ) = . 1935 Problem 6 . 05 ⋅ . 7 + . 95 ⋅ . 07 = 10 . 15% 3...
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## This note was uploaded on 04/05/2012 for the course STATS V3100018.0 taught by Professor Giuseppearbia during the Spring '12 term at NYU.

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Solutions to Problem Set 5 - b. P ( one and only one 3 ∩...

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