math414 hw5

# math414 hw5 - Stat/Math 414 Solutions 5 1.4-11 a Denote...

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Stat/Math 414 Solutions # 5 1.4-11 . a) Denote this event by A . Let S be the space of all unordered samples without replacement of size 3 from 12 bulbs. We denote by N the size of the outcome space S, N ( A ) the number of all favorable outcomes. N = ± 12 3 , N ( A ) = ± 3 3 = 1; P ( A ) = N ( A ) N = 1 220 . b) Denote this event by A . B=“There were 2 defective bulbs in the ﬁrst 4 tests”. C=“The 5th tested bulb is defective”. A = B C, P ( A ) = P ( B C ) = P ( C | B ) P ( B ) P ( B ) = ( 3 2 )( 9 2 ) ( 12 4 ) = 24 110 ; P ( C | B ) = 3 - 2 12 - 4 = 1 8 . Hence P ( A ) = 24 110 · 1 8 = 3 110 . c) Let this event be A . B=“There were 2 defective bulbs in the ﬁrst 9 tests”. C=“The 10th tested bulb is defective. A = B C, P ( B ) = ( 3 2 )( 9 9 - 2 ) ( 12 9 ) = 27 55 ; P ( C | B ) = 1 3 . P ( A ) = 27 55 · 1 3 = 9 55 . 1.4-14 . N = 52 6 ; N ( A ) = 52 · 51 · 50 · 49 · 48 · 47. P

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math414 hw5 - Stat/Math 414 Solutions 5 1.4-11 a Denote...

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