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 February 24th, 2013, 10:47 AM #1 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Probability and Arrangements 1. The Allens wanted to have 5 children. If they had all 5 children, what is the probability that at least two of them are boys?[list:1u9cfgh0] A. 3/16 B. 13/16 C. 11/16 D. 5/16 2. Mrs. Crawford has an algebra, a geometry book, a trigonometry book, and a calculus book to put on her bookshelf. If the geometry book is not on either end and the calculus book is to the left of the trig book, in how many ways can she arrange the 4 books?A. 3 B. 6 C. 9 D. 12 [/list:u:1u9cfgh0] I figure that 1. is 13/16 (I did a tree diagram), and I figure that 2. is 3 (I wrote it all out). My questions are: a) Are these the correct answers? b) How do I do derive these answers elegantly? (I.e., from basic principles of counting, combinatorics, and probability.)
 February 24th, 2013, 11:06 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Probability and Arrangements 1.) I would look at the number of way for the Allens to have 5 or 4 girls. There is 1 way to have 5 girls, and 5 ways to have 4 girls, for a total of 6. Since there are 32 outcomes, then the number of ways to have at least 2 boys is: $P(X)=1-\frac{6}{32}=1-\frac{3}{16}=\frac{13}{16}$
February 24th, 2013, 11:12 AM   #3
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Re: Probability and Arrangements

Quote:
 Originally Posted by MarkFL 1.) I would look at the number of way for the Allens to have 5 or 4 girls. There is 1 way to have 5 girls, and 5 ways to have 4 girls, for a total of 6. Since there are 32 outcomes, then the number of ways to have at least 2 boys is: $P(X)=1-\frac{6}{32}=1-\frac{3}{16}=\frac{13}{16}$
That's a very nice way of looking at it, Mark! I will try that whenever I encounter these problems. I would predict that, in general, it greatly simplifies the calculations.

February 24th, 2013, 11:46 AM   #4
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Re: Probability and Arrangements

Quote:
Originally Posted by CherryPi
Quote:
 Originally Posted by MarkFL 1.) I would look at the number of way for the Allens to have 5 or 4 girls. There is 1 way to have 5 girls, and 5 ways to have 4 girls, for a total of 6. Since there are 32 outcomes, then the number of ways to have at least 2 boys is: $P(X)=1-\frac{6}{32}=1-\frac{3}{16}=\frac{13}{16}$
That's a very nice way of looking at it, Mark! I will try that whenever I encounter these problems. I would predict that, in general, it greatly simplifies the calculations.
I am confident that you will find your prediction borne out!

February 24th, 2013, 04:02 PM   #5
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Re: Probability and Arrangements

Quote:
 Originally Posted by CherryPi 2. Mrs. Crawford has an algebra, a geometry book, a trigonometry book, and a calculus book to put on her bookshelf. If the geometry book is not on either end and the calculus book is to the left of the trig book, in how many ways can she arrange the 4 books? A. 3 B. 6 C. 9 D. 12
I think the answer is 6. Temporarily ignore the geometry book. Since calculus is to the left of trig, then the calc, trig, and algebra books must be one of these three:

CAT
CTA
ACT

For each of these there are two places to insert the geometry book. Thus 3 x 2 = 6. Here they are:

CGAT CAGT
CGTA CTGA
AGCT ACGT

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