Question

Suppose that a nonnegative integer solution to the equation u + v + w + x + y = 12, thought of as a 5-tuple (u,v,w,x,y) is chosen at random (each one being equally likely to be chosen). If one solution is chosen at random, what is the probability that u = 1?

Answer 1

Answer:0.2

Step-by-step explanation:

Given

$u+v+w+x+y=12$

total solution for $u+v+w+x+y =12$ is given by $^{n+r-1}C_{r-1}$

here $n=12\ r=5$

i.e. $^{12+5-1}C_{5-1}=^{16}C_4$

For $u=1$

$1+v+w+x+y=12$

$v+w+x+y=11$

no of solutions for this equation is

$^{11+4-1}C_{4-1}=^{14}C_3$

if one solution is chosen at random Probability that u=1

$=\frac{^{14}C_3}{^{16}C_4}$

$=\frac{364}{1820}=0.2$