Question

# Three randomly chosen natural numbers x,y and z satisfy x+y+z=10, then

A
Probability that z is even is 12
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B
Probability that z is even is 49
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C
Probability that z is odd is 59
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D
Probability that x is even is 512
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Solution

## The correct options are B Probability that z is even is 49 C Probability that z is odd is 59x+y+z=10, x,y,z∈N Put x=a+1, y=b+1, z=c+1, a,b,c∈W a+b+c=7 Total solutions = 7+3−1C3−1=36 For z to be even, c must be an odd number. If c=1⇒a+b=6. Number of solutions = 6+2−1C2−1=7 If c=3⇒a+b=4. Number of solutions = 4+2−1C2−1=5 If c=5⇒a+b=6. Number of solutions = 2+2−1C2−1=3 If c=7⇒a+b=6. Number of solutions = 0+2−1C2−1=1 Number of solution when c is odd = 16 Probability that z is even =1636=49 Probability that z is odd =2036=59

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