Homework_2_Solution

Homework_2_Solution - Homework 2 Solution 1.2 -12 (4...

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Homework 2 Solution 1.2 -12 (4 points) a. P({x: 0 ≤ x ≤ 1/3}) = 1/3 b. P({x: 1/3 ≤ x ≤ 1}) = 2/3 c. P({x: x = 1/3}) = 0 d. P({x: 1/2 < x < 5}) = P({x: 1/2 < x ≤ 1}) = 1/2 1.2- 17 (8 points) P(S) = P(A 1 A 2 A m ) = P(A 1 ) + P(A 2 ) + … + P(A m ) = 1 since A 1 , A 2 , …, A m are mutually exclusive and exhaustive. a. If P(A 1 ) = P(A 2 ) = … = P(A m ), then P(A 1 ) + P(A 2 ) + … + P(A m ) = mP(A i ) = 1 for all i = 1, 2, …, m. Therefore, P(A i ) = 1/m for all i = 1, 2, …, m. b. P(A) = P(A 1 A 2 A h ) = P(A 1 ) + P(A 2 ) + … + P(A h ) since A 1 , A 2 , …, A h are mutually exclusive. Use the result from (a) to replace all P(A i ) by 1/m we have P(A) = h/m. 1.3 – 2 (3 points) For the first position we have 4 options. For the second position we have 3 options since one of four flags has been selected. For the third position we have 2 options. Therefore, the total number of different signals can be made is (4)(3)(2) = 24.
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This note was uploaded on 09/16/2008 for the course STAT 414 taught by Professor Senturk,damla during the Fall '07 term at Penn State.

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Homework_2_Solution - Homework 2 Solution 1.2 -12 (4...

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