Mary is shopping for houses in the neighborhood. 20 of the houses are between $60,000-$70,000; 30 of them are $80,000-$90,000; 20 of the houses are between $91,000-$110,000; and 30 of the houses are above $120,000. If Mary is debating between two houses, one that is $80,000-$90,000 and another that is $91,000-$110,000, what are the number of ways that Mary can select houses?
Mary is shopping for houses in the neighborhood. 20 of the houses are between $60,000-$70,000; 30 of them are $80,000-$90,000; 20 of the houses are between $91,000-$110,000; and 30 of the houses are above $120,000. If Mary is debating between two houses, one that is $80,000-$90,000 and another that is $91,000-$110,000, what are the number of ways that Mary can select houses?
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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Mary is shopping for houses in the neighborhood. 20 of the houses are between $60,000-$70,000; 30 of them are $80,000-$90,000; 20 of the houses are between $91,000-$110,000; and 30 of the houses are above $120,000. If Mary is debating between two houses, one that is $80,000-$90,000 and another that is $91,000-$110,000, what are the number of ways that Mary can select houses?
Expert Solution
Step 1
Given data
Price Range |
No of Houses |
$60,000 -$70,000 |
20 |
$80,000 -$90,000 |
30 |
$91,000 -$110,000 |
20 |
Above $120000 |
30 |
No of houses Mary can select in the range of $80,000 -$90,000 = 30
No of houses Mary can select in the range of $90,000 -$110,000 = 20
Total no of ways Mary can select the Houses = 30 x 20 = 600
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