B) 5/21?

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Suppose that a solution $(X,Y,Z)$ of the equation $X+Y+Z=20,$with $X,Y$ and $Z$ non-negative integers,

is chosen at random.What is the probability that $X$ is divisible by $5?$

$A)\frac{1}{4}$

$B)\frac{5}{21}$

$C)\frac{2}{7}$

$D)\frac{3}{20}$

is chosen at random.What is the probability that $X$ is divisible by $5?$

$A)\frac{1}{4}$

$B)\frac{5}{21}$

$C)\frac{2}{7}$

$D)\frac{3}{20}$

+4

the procedure will be in this way:

Number of non-negative integral solutions = ^{22}C_{2 }

For finding X for divisible by 5:

X = 0 then Y + Z = 20 Number of non negative integral solutions = ^{21}C_{1 }= 21

X =5 then Y+ Z = 15 Number of non negative integral solutions = ^{16}C_{1 } = 16

X=10 then Y + Z = 10 Number of non negative integral solutions =^{11}C_{1 } = 11

X =15 then Y + Z = 5 Number of non negative integral solutions = ^{6}C_{1 }= 6

X =20 then Y + Z =0 Number of non negative integral solutions = ^{1}C_{1} = 1

Sum of all these values = 21 + 16 + 11 + 6 + 1 = 55

Required probability = 55 / ^{22}C_{2} = 5/21