# soln2 - S2-1EEL 5544 HW 2 Solutions Fall...

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Unformatted text preview: S2-1EEL 5544 HW 2 Solutions, Fall 2008CombinatoricsSS-1.(a) With order and with repeats, the number of possible “combinations” (this is a dangerousword to use here when talking about probabilities) is 603=216,000.(b) With order and without repeats, the number of “combinations” is 60·59·58=205320.(c) Without order and without repeats, the number of combinations isC603=34220.SS-2.S2-21.3^--lC S V * 5 v x^u v^S2-3^ . ^ J w - W V » A s ^ t » S ^ " 5 "SS-3. This problem is easier to solve with ordering, since the two cards are different.Thenblackjack consists of an Ace, of which there are 4, paired with one of the 16 cards thathave the value 10 (the 10, Jack, Queen, and King). The blackjack can then occur in twoorders: T-A or A-T, where T represents one of the cards with value 10. The number ofordered pairs of cards is 52·51, so the probability of blackjack isP(blacjack) =(4)(16)(2)(52)(51)≈.0483S2-4SS-4. This problem is easier to solve if we consider the ordered outcomes as our sample space.Otherwise, the outcomes will not have equal probability.Make sure that you understandwhy!Unlike drawing cards, repeatedly rolling dice allows replacement, so the total numberof outcomes when rolling 5 dice (or rolling a single die 5 times) is 65.(a) The number of ordered ways to make “no two alike” is as follows: There are 6 choicesfor the first die. Then the second die can be any of the remaining values, and so on. So,the probability of no two alike is6·5·4·3·265≈.0926(b) The number of ordered ways to make a pair is as follows: The number of differentdice values is now reduced to 6·5·4·3, because two dice have the same value. Now,these represent the ordered way in which the values show up, but one of the 3 values isrepeated. We need to count the number of ways in which that value can appear twice,which is52.Thus the probability of a pair is6·5·4·3(52)65≈.4630(c) The number of ordered ways to make two pair is as follows: The number of differentdice values is now reduced to 6·5·4, because there are two pairs of dice with the samevalue. Now, these represent the ordered way in which the values show up, but we mustcount the number of ways that the two pairs can be distributed among the 5 dice. Thus,from above, the first pair can appear in...
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## This note was uploaded on 01/29/2012 for the course ECE 101 taught by Professor Wang during the Spring '11 term at Iowa State.

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soln2 - S2-1EEL 5544 HW 2 Solutions Fall...

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