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# practice_exam1_solution[1] (1) - 036 054 = 0 667 1 4...

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STAT400 Discussion sections Midterm 1 Practice Problems - Solution 1. a) Since A and B are disjoint, P ( A B ) = , and thus, P ( A B ) = P ( A ) + P ( B ). Therefore P ( A ) = P ( A B ) - P ( B ) = 0 . 5 - (1 - 0 . 7) = 0 . 2 . b) Note that P ( A B ) + P ( A B 0 ) + P ( A 0 B ) + P ( A 0 B 0 ) = 1. Thus, P ( A B 0 ) + P ( A 0 B ) + P ( A B ) = 1 - P ( A 0 B 0 ) = 1 - P ( A 0 ) P ( B 0 ) ( A and B are independent) = 1 - (1 - 0 . 6) × (1 - 0 . 7) = 1 - 0 . 4 × 0 . 3 = 0 . 88 c) Recall De Morgan’s law, ( A B ) 0 = A 0 B 0 . Then, P ( A 0 B 0 ) = 1 - P ( A B ) = 1 - { P ( A ) + P ( B ) - P ( A B ) } = 1 - (0 . 22 + 0 . 25 - 0 . 11) = 0 . 64 2. Let A be the event that we have one ace, two faces, and two other cards. We need to choose 1 ace out of 4 aces, two faces out of 3 × 4 faces, and two cards out of 52 - 4 - 12 remaining cards. Thus, P ( A ) = N ( A ) N ( S ) = 4 1 12 2 36 2 52 5 = 0 . 06399 3. Define the following events; M = { Male } , F = { Female } , U = { Unemployed } . a) P ( M U ) = P ( U | M ) P ( M ) = 0 . 06 × 0 . 6 = 0 . 036 b) P ( U ) = P ( M U )+ P ( F U ) = P ( U | M ) P ( M )+ P ( U | F ) P ( F ) = 0 . 045 × 0 . 4+0 . 06 × 0 . 6 = 0 . 054 c) P ( M | U ) = P ( M U ) P ( U ) = 0 .

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Unformatted text preview: . 036 . 054 = 0 . 667 1 4. Premium = Expected Loss = \$40 , 000 × . 001 + \$20 , 000 × . 005 = \$40 + \$100 = \$140 5. Note that X follows b (25 , . 1). Refer to the CDF table for binomial distribution on page 579. a) P ( X ≤ 3) = 0 . 7636 b) P ( X = 3) = P ( X ≤ 3)-P ( X ≤ 2) = 0 . 7636-. 5371 = 0 . 2265 c) P ( X ≥ 3) = 1-P ( X ≤ 2) = 1-. 5371 = 0 . 4629 d) P (2 ≤ X ≤ 4) = P ( X ≤ 4)-P ( X ≤ 1) = 0 . 9020-. 2712 = 0 . 6308 6. a) (0 . 85) 3 (0 . 15) b) (0 . 85) 6 (0 . 15) c) ± 15 4 ² (0 . 15) 4 (0 . 85) 11 2...
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