Mixed Balls
Ten balls labelled are put into an urn and thoroughly mixed. Then, one by one, each ball is taken from the urn at random and ordered in a line. The probability that no two of the prime-numbered balls are placed next to each other may be expressed in the form , where and are positive coprime integers. What is ?
The answer is 7.
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1 solution
Look, solution to this problem is quite simple. Total no of ways to arrange the balls is 10!(! denotes factorial), and it is our sample space. Now no of favourable events is calculated as below:- -2-3-5-7- ('-' denotes spaces occupied by integers). Let the five spaces be a,b,c,d,e respectively. Out of them a,e can be 0 but b,c,d have to be at least 1. Thus you may form the equation, a+(b-1)+(c-1)+(d-1)+e=3 and no of integral solutions comes out to be 35(via fundamental principle of counting). Thus, our required probability is (35x4!x6!)/(10!).(4!&6! to count permutations of prime and non-prime no's). This comes out as 1/6. Thus our answer is 1+6=7.