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 3655 611 12  511

Solution

## The correct option is B 611 x+y+z=10 Total no. of solutions = 10+3−1C3−1=66 Let z=2n⇒x+y=10−2n; n∈{0,1,2,3,4,5} Favourable number of solutions= (10−2n)+2−1C2−1=11−2n ∴Total favourable solutions =5∑n=0(11−2n)=36 Required probability =3666=611

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