Q. 1)A rod of length 10cm is broken into 3 parts such that the parts form a triangle and are an integral multiple of 1cm.Find its probability. 2)out of n persons sitting in a round table, three, A, B and C are chosen at random. Find the chance that no 2 of these 3 are sitting together.plz explain the reasons.

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).

Approved

Very easy sum,dude :)