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Solution 4

The document presents solutions to various combinatorial problems, including permutations and combinations. It covers calculations for selecting items with and without regard to order, as well as probabilities of specific outcomes. Additionally, it discusses the relationship between subsets of different sizes.

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24 views2 pages

Solution 4

The document presents solutions to various combinatorial problems, including permutations and combinations. It covers calculations for selecting items with and without regard to order, as well as probabilities of specific outcomes. Additionally, it discusses the relationship between subsets of different sizes.

Uploaded by

tonytinyiko
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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STA01A1 Solution 4, University of Johannesburg 1

SOLUTION 4
30.
a. Because order is important, we’ll use P8,3 = 8(7)(6) = 336.

b. Order doesn’t matter here, so we use C30,6 = 593 775

 8  10  12 
c. From each group we choose 2:   •   •   = 83 160
 2  2   2 
83 160
d. The numerator comes from part c and the denominator from part b: = 0.14
593 775
e. We use the same denominator as in part d. We can have all zinfandel, all merlot, or all cabernet,
 8  10  12 
  +   +  
so P(all same) = P(all z) + P(all m) + P(all c) =
 6   6   6  = 1162 = 0.002
 30  593 775
 
6
31.
a. (n1)(n2) = (9)(5) = 45

b. (n1)(n2)(n3) = (9)(5)(32) = 1440, so such a policy could be carried out for 1440 successive nights,
or approximately 4 years, without repeating exactly the same program.

35.
STA01A1 Solution 4, University of Johannesburg 2

 5
36. There are 10 possible outcomes --   ways to select the positions for B’s votes: BBAAA, BABAA,
 2
BAABA, BAAAB, ABBAA, ABABA, ABAAB, AABBA, AABAB, and AAABB. Only the last two
have A ahead of B throughout the vote count. Since the outcomes are equally likely, the desired
probability is 102 = .20 .

39.
 6  9 
  
a. P(selecting 2 - 75 watt bulbs) =
 2  1  = 15  9 = .2967
15  455
 
3

 4   5  6 
  +   +  
b. P(all three are the same) =
 3   3   3  = 4 + 10 + 20 = .0747
15  455
 
3

 4  5  6 
   
c.
 1  1  1  = 120 = .2637
15  455
 
3

44.
 n n! n!  n 
  = = =  
 k  k!(n − k )! (n − k )!k!  n − k 
The number of subsets of size k = the number of subsets of size n-k, because to each subset of size k
there corresponds exactly one subset of size n-k (the n-k objects not in the subset of size k).

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