sir,how could this be solved,
there are 10 pairs of socks,in a cupboard from which 4 individual socks are picked at random.the probablity that there is at least one pair is.
The sample space would be 20C4. The number of ways of picking 4 socks with no matching pair would be 20 x 18 x 16 x 14 / 4!. Find the probability accordingly. Then, the probability of at least one pair is 1 – the above.
It can be looked at as the same logic as integration – area under the curve. At 5:00 the probability is 1/6, at 5:10 it is 1/3 and in between it increases linearly. So the average probability at any time in that interval will be just the average of the two extremes. Had it not been a linear increase then it would have been a different (and more complicated) story.
CAT-holics 
sir,how could this be solved,
there are 10 pairs of socks,in a cupboard from which 4 individual socks are picked at random.the probablity that there is at least one pair is.
The sample space would be 20C4. The number of ways of picking 4 socks with no matching pair would be 20 x 18 x 16 x 14 / 4!. Find the probability accordingly. Then, the probability of at least one pair is 1 – the above.
regards
J
Why do we take average of both the probabilities to calculate probability of first and the last interval?
It can be looked at as the same logic as integration – area under the curve. At 5:00 the probability is 1/6, at 5:10 it is 1/3 and in between it increases linearly. So the average probability at any time in that interval will be just the average of the two extremes. Had it not been a linear increase then it would have been a different (and more complicated) story.
regards
J
Got it..thanks!!