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100Ahw2key

# 100Ahw2key - B be the event that an empty door is opened p...

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STAT 100 Homework II Answer Key Problem 1: Let X and Y be the two numbers respectively. p ( X > 3) = 1 / 3 . p ( X > 3 | X + Y > 9) = 1. p ( X = k || X - Y | > 2) = p ( X = k, | X - Y | > 2) /p ( | X - Y | > 2). p ( | X - Y | > 2) = 12 / 36 = 1 / 3. p ( X = 1 , | X - Y | > 2) = 3 / 36 = 1 / 12. So p ( X = 1 || X - Y | > 2) = 1 / 4. Similarly, p ( X = 2 || X - Y | > 2) = 1 / 6. p ( X = 3 || X - Y | > 2) = 1 / 12. p ( X = 4 || X - Y | > 2) = 1 / 12. p ( X = 5 || X - Y | > 2) = 1 / 6. p ( X = 6 || X - Y | > 2) = 1 / 4. Problem 2: p ( X > . 5 | X + Y > 1 . 5) = 1. p ( Y < . 3 || X - Y | > . 5) = p ( Y < . 3 , | X - Y | > . 5) /p ( | X - Y | > . 5) = . 5 × ( . 5 × . 5 - . 2 × . 2) /. 5 × . 5 = . 42. Problem 3: p ( D | +) = p ( D, +) /p (+) = p ( D ) p (+ | D ) / ( p ( D ) p (+ | D )+ p ( N ) p (+ | N )) = . 1 × . 99 / ( . 1 × . 99 + . 99 × . 1) = . 5. Problem 4: p ( degree ) = p ( under 40) p ( degree | under 40) + p ( above 40) p ( degree | above 40) = 1 / 2 × 1 / 3 + 1 / 2 × 2 / 3 = 1 / 2. p ( under 40 | degree ) = p ( under 40) p ( degree | under 40) /p ( degree ) = (1 / 2 × 2 / 3) / (1 / 2) = 2 / 3. Problem 5: Let X be the initial state of a random person. Let Y be the state of this person after immigration. p ( Y = j ) = 5 i =1 p ( X = i ) p ( Y = j | X = i ). p ( X = i | Y = j ) = p ( X = i ) p ( Y = j | X = i ) /p ( Y = j ). Problem 6: p ( second = red ) = p ( first = red ) p ( second = red | first = red ) + p ( first = blue ) p ( second = red | first = blue ) = [ r/ ( r + b )] × [( r +1) / ( r + b +1)]+[ b/ ( r + b )] × [ r/ ( r + b +1)] = r/ ( r + b ). Problem 7: Let A be the event that there is gold behind the picked door. Let
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Unformatted text preview: B be the event that an empty door is opened. p ( A ) = 1 / 3. If the host knows where the gold is and open the empty door intentionally, then p ( B | A ) = 1, and p ( B ) = 1. So p ( A | B ) = p ( A ) p ( B | A ) /p ( B ) = 1 / 3. So one should switch. If the host does not know where the gold is, and he happens to open an empty door, then p ( B | A ) = 1, and p ( B | ¯ A ) = 1 / 2. So p ( B ) = p ( A ) p ( B | A )+ p ( ¯ A ) p ( B | ¯ A ) = 1 / 3 × 1+ 2 / 3 × 1 / 2 = 2 / 3. p ( A | B ) = p ( A ) p ( B | A ) /p ( B ) = 1 / 3 × 1 / (2 / 3) = 1 / 2, so there is no need to switch. 1...
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