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Question

Three randomly chosen non-negative integers x, y and z are found to satisfy the equation x+y+z=10. Then the probability that z is even, is 
  1. 3655 
  2. 611 
  3. 12  
  4. 511  


Solution

The correct option is B 611 
x+y+z=10
Total no. of solutions = 10+31C31=66
Let z=2nx+y=102n; n{0,1,2,3,4,5}
Favourable number of solutions= (102n)+21C21=112n
Total favourable solutions
=5n=0(112n)=36
Required probability
=3666=611

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