Module 3 written - are dealt a “pocket pair” that is...

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Derrick Berry Written Assignment 3 The probability that you are dealt pocket aces is 1/221, or 0.00452 to three significant digits. If you studied either section 4.5 and 4.6 or section 4.8, verify that probability. Ace #1: is a chance of 4 out of 52 or P(Ace) = 4/52 = 0.07690 Ace # 2: is a chance of 3 out of 51 or P(Ace 2) = 3/51 = 0.05882 P(Ace and Ace 2) = P(Ace) x P(Ace 2) = 0.07690 x 0.05882 = .00452 Using the result from part (a), obtain the probability that you are dealt “pocket kings”. King #1: is a chance of 4 out of 52 or P(King) = 4/52 = 0.07690 King # 2: is a chance of 3 out of 51 or P(King 2) = 3/51 = 0.05882 P(King and King2) = P(King) x P(King 2) = 0.07690 x 0.05882 = .00452 Using the result from part (a) and your analysis in part (b), find the probability that you
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Unformatted text preview: are dealt a “pocket pair”, that is, two cards of the same denomination. (4/52)(3/51)(13)=0.05882 Contains at least 1 card of your denomination (Hint: Complementation rule.) (2/50)(48/49)(47/48) + (2/50)(1/49)(48/48)=0.03918 Gives you “trips” that is , contains exactly 1 card of your denomination and 2 other unpaired cards. (2/50)(48/49)(44/48)=0.03592 Gives you “quads” that is, contains 2 cards of you denomination . (2/50)(1/49)(48/48)=0.00082 Gives you a “boat” that is, contains 1 card of your denomination and 2 cards of another denomination. (2/50)(48/49)(3/48)=0.00245...
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