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# 7146330 - 5 c Prob(David is not dealt a K or a Q To...

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Unformatted text preview: 5 c) Prob(David is not dealt a K or a Q) To calculate this, count the number of ways David is dealt a hand from the remaining cards (after removing the K's and Q's 16 remain). This is equal to (16,5). Divide this by the total number of ways to select 5 cards out of 24 cards. 5 d) By de Morgan's law (AUB)c = Ac Bc A is the event that peggy is dealt no king. B is the event that peggy is dealt no queen. We had calculated the probability that a player was not dealt a K or a Q in part c of question ve. Using that and de Morgan's law, we get the answer. 5 e) Here, I made a small error and I will recalculate. P (K Q) P(K|Q)= P (Q) P(K Q)= (4,1)(4,1)(22,3) The numerator is the number of ways of choosing (24,5) one king and one queen from the deck and then arbitrarily choosing the remaining cards. P(Q)=1 − (20,5) (24,5) (20,5) Probability of not getting a queen = (24,5) where the numerator is no of ways of selecting 5 cards from a suit of 20(without queens) so (4,1)(4,1)(22,3) P(K|Q)= (24,,5) (20 5) (24,5) 5 f) This question is a little unclear because in part b we are asked to calculate probability of getting a queen. Whereas, in part c we are asked to calculate probability of getting a K given that we have a queen. If we reinterpret part b as calculating probability of getting a K (it makes no dierence to the calculation), we can say that answer in part e is greater than the answer in part b. This is because we are given information the distribution, our probability space has become smaller. Correspondingly, the probabilities of events increases to keep the sum equal to 1. 1 ...
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