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Hi Ashmita,
1. Let x1,x2,x3 be the lengths into which the 10cm rod is broken into.
(Say x1≤x2≤x3)..... also clearly x1+x2+x3 = 10.
So lets see all possibilities.
1,1,8 2,2,6 3,3,4
1,2,7 2,3,5
1,3,6 2,4,4
1,4,5
So out of these, we have to decide which can form a triangle.
For a triangle to be formed, the sum of the sides of any two sides should be more than the third side (Triangle Property)
So only (2,4,4) and (3,3,4) are possible
So 2/8 is the probability = 1/4 = 0.25
2. For this question, the total number of ways of selecting 3 ppl from "n" would be nC3.
Now we have to find the favourable cases, where we have to select according to the given condition....
Consider..... (n-3) peopple in a straight line, like this
x x x x x x x ...... x ------------(n-3 crosses)
Now there are n-2 gaps here, where 3 people can be placed in n-2C3 ways (so that they are not adjacent)
Now if we place people in the 1st place before the 1st x and in the last place after the last x, in a circle they will be adjacent (so this case has to be removed)
The no of arrangements in which 1 person is in the 1st place, and another person in the last place is n-4.
So the favourable cases is n-2C3 - (n-4)
The probability would be {n-2C3 - (n-4)}/nC3.
Regards,
Ashwin (IIT Madras).
Very easy sum,dude :)
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