Solution_to_Sample_Midterm_1

# Solution_to_Sample_Midterm_1 - STAT/MATH 418 1st Sample...

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Unformatted text preview: STAT/MATH 418 1st Sample Midterm Solution Solution to Sample Midterm 1 1. (a) 10! 5!5! (b) 3! 4! 3! 2! 2!2! 2! 2. 15 terms x y z Num. of terms 3!/ 2! = 3 3! 3!/ 2! = 3 3!/ 2! = 3 4 3 2 2 0 1 1 2 0 0 1 0 4 1 2 1 2 ( x ) ( 2 y ) ( z ) = 48 xy z 1, 2,1 c c 3. We know from the problem P ( M ) = 0.6, P ( J ) = 0.4 and P ( M J ) = 0.2. We want to know P ( M y J ) , that is P ( M U J ) = P( M ) + P ( J ) - P ( M I J ) = 0.6 + 0.4 - P ( M I J ) Since P ( M U J ) = 0.8 from P ( M c I J c ) = P ( M U J ) = 0.2, 0.8 = 0.6 + 0.4 - P ( M I J ) c P ( M I J ) = 0.2 4. P ( A ) = 0.3, P ( B) = 0.4 and P (C ) = P ( A U B ) = 0.2 Therefore, P ( A U B ) = P ( A ) + P ( B ) - P ( A U B ) = 0.3 + 0.4 - 0.2 = 0.5 10 5. (i) 3! 3 10 8 (ii) There are three cases ; choose A and no B, B and no A, and no A and no B. So, 3! . - 1 3 50 4 6. P ( at least one statistician ) = 1 - P ( no statiatician ) = 1 - 25 25 0 4 ...
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