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By Sheldon Ross

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A) 1/16 (b) 1/16 (c) The only way in which the pattern H, H, H, H can occur first is for the first 4 flips to all be heads, for once a tail appears it follows that a tail will precede the first run of 4 heads (and so T, H, H, H will appear first). Hence, the probability that T, H, H, H occurs first is 15/16. 60. From the information of the problem we can conclude that both of Smith’s parents have one blue and one brown eyed gene. Note that at birth, Smith was equally likely to receive either a blue gene or a brown gene from each parent.

Hence, P(E b F) = 77. P( E ) . P( E ) + P( F ) (a) This is equal to the conditional probability that the first trial results in outcome 1 (F1) given that it results in either 1 or 2, giving the result 1/2. More formally, with L3 being the event that outcome 3 is the last to occur P(F1L3) = P( L3 F1 ) P ( F1 ) P ( L3 ) = (1/ 2)(1/ 3) = 1/ 2 1/ 3 (b) With S1 being the event that the second trial results in outcome 1, we have P(F1S1L3) = 78. P( L3 F1 S1 ) P( F1S1 ) P( L3 ) = (1/ 2)(1/ 9) = 1/ 6 1/ 3 (a) Because there will be 4 games if each player wins one of the first two games and then one of them wins the next two, P(4 games) = 2p(1 − p)[p2 + (1 − p)2].

P(BjNi) = P( N i B j ) P( B j ) P( N i Bi ) P( Bi ) + P( N i Bic ) P( Bic ) Pj if j ≠ i (1 − α i ) Pi + 1 − Pi (1 − α i ) Pi if j = i = (1 − α i ) Pi + 1 − Pi = 5. None are true. 6. n  n  P ∪ Ei  = 1 − P ∩ Eic  = 1 − 1  1  7. n ∏[1 − P( E )] i 1 (a) They will all be white if the last ball withdrawn from the urn (when all balls are withdrawn) is white. As it is equally likely to by any of the n + m balls the result follows. g g b P ( RBG G last) = . r +b+ g r +b+ g r +b bg b g Hence, the answer is .

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