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

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

`=(^(14)C_3)/(^(16)C_4)`

`=(364)/(1820)=0.2`