challset2sol - 1 Mat 344F challenge set #2 Solutions 1. Put...

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1 Mat 344F challenge set #2 Solutions 1. Put two balls into box 1, one ball into box 2 and three balls into box 3. The remaining 4 balls can now be distributed in any way among the three remaining boxes. This can be done in C(4+3-1,4)=C(6,4)=15 ways. Please note that the question should have said that box 1 has at least two balls. 2. There are mn entries in the m x n matrix, two choices for each, and the choices are independent of one another, hence by the multiplication principle, there are 2 mn matrices that have 0,1 entries. If we allow the entries to be 0,1,2,3 then each of the mn entries has four choices which are independent of one another. The multiplication principle tells us that there are 4 mn such matrices. 3. Let A be the event,” exactly one pair of shoes is selected”. Let B be the event “exactly two pairs of shoes are selected. Let S be the event “ four shoes are selected from 20 shoes with order being irrelevant”. We want to calculate P(at least one pair of shoes is selected) = P(A or B) = P(A) +P(B) since events A and B are disjoint. We will use the probability formula P(A) = n(A) n(B) . The number of ways to select exactly one pair of shoes = n(A) = C(10,1) x (C(18,2)- 9). C(10,1) counts the number of ways to pick one pair of shoes. There are 18 shoes remaining, pick two shoes. However we don’t want to pick another pair so we must subtract the 9 remaining pairs of shoes. The number of ways of picking two pairs of shoes = n(B) = C(10,2). Using these results, it follows that P(selecting at least one pair of shoes) = C C ( , ) ( , ) . 10 1 20 4 x(C(18,2) - 9) C(10,2) C(20,4) + Another way to approach this problem is to first calculate the probability that no pair of shoes is selected and then subtract this from one. This will yield the probability that at least one pair of shoes is selected. The number of ways to select no pair of shoes is C(10,4) x C(2,1) x C(2,1) x C(2,1) xC(2,1). C(10,4) counts the number of ways to select four shoes out of ten with order being irrelevant. Once the pairs have been selected, then out of each pair we can select one shoe, and this
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2 can be done in (C(2,1)) 4 ways. The probability of selecting at least one pair of shoes is 1 20 4 - C(10,4)x2 4 C ( , ) . 4. Fix the six individuals in some order and suppose we have two Monday, Wednesday and Friday labels. We will place one label on each person. The end result will be two accidents with Monday labels, two accidents with Wednesday and two accidents with Friday labels. This is like having six distinct items, two each of three kinds, and we want all the possible arrangements. The number of such arrangements is 6! 2 2 2 ! ! ! . You can think of this as the number of ways to arrange the letters MM WW FF (i.e. permutations with repetition). Using the permutation with repetitions formula, the number of arrangements is 6! 2 2 2
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This note was uploaded on 03/04/2012 for the course BIO 510 taught by Professor Miller during the Fall '06 term at Carnegie Mellon.

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challset2sol - 1 Mat 344F challenge set #2 Solutions 1. Put...

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