Unformatted text preview: = 2 } = 5 10 × 5 9 P { X = 3 } = 5 10 × 4 9 × 5 8 P { X = 4 } = 5 10 × 4 9 × 3 8 × 5 7 P { X = 5 } = 5 10 × 4 9 × 3 8 × 2 7 × 5 6 P { X = 6 } = 5 10 × 4 9 × 3 8 × 2 7 × 1 6 × 5 5 Alternative solution. Consider the event X ≥ i . It is equivalent to saying that the top i1 individuals are men. Hence, P { X ≥ i } = ( 5 i1 ) ( 10 i1 ) Now, P { X = i } = P { X ≥ i } P { X ≥ i + 1 } = ( 5 i1 ) ( 10 i1 )( 5 i ) ( 10 i ) 1...
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This note was uploaded on 04/02/2008 for the course MATH 425 taught by Professor Buckingham during the Winter '08 term at University of Michigan.
 Winter '08
 Buckingham
 Math, Probability

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