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Source: — Problem Solving |
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by cramya » Sun Dec 28, 2008 10:49 am
p(A atleast 5) = 2/3
p(A not atleast 5) = 1/3

p(B atleast 5) = 1/4
p(B not atleast 5) = 3/4



p(atleast one) = 1 - p(none)

= 1- (1/3) * (3/4)

= 9/12
=3/4

Hope this is right since prob has never been my cup of tea and trying my best to make it so......
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by parallel_chase » Sun Dec 28, 2008 11:22 am
probability of roads from A-B at least 5 miles = 2/3
probability of roads from A-B NOT at least 5 miles = 1- 2/3 = 1/3

probability of roads from B-C at least 5 miles = 1/4
probability of roads from B-C NOT at least 5 miles = 1- 1/4 = 3/4

There are 3 cases - at least 1 road selected should be at least 5 miles long

CASE 1=probability of selecting road 5 miles long A-B * probability of NOT selecting road 5 miles long B-C
2/3 * 3/4 = 6/12

CASE 2=probability of NOT selecting road 5 miles long A-B * probability of selecting road 5 miles long B-C
1/3 * 1/4 = 1/12

CASE 3=probability of selecting road 5 miles long A-B * probability of selecting road 5 miles long B-C
2/3 * 1/4 = 2/12

Total probability = 6/12 + 1/12 + 2/12 = 9/12 = 3/4
No rest for the Wicked....
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by rajataga » Sun Dec 28, 2008 11:33 am
right you are :)
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