Respuesta :
Answer:0.2
Step-by-step explanation:
Given
[tex]u+v+w+x+y=12[/tex]
total solution for [tex]u+v+w+x+y =12[/tex] is given by [tex]^{n+r-1}C_{r-1}[/tex]
here [tex]n=12\ r=5[/tex]
i.e. [tex]^{12+5-1}C_{5-1}=^{16}C_4[/tex]
For [tex]u=1[/tex]
[tex]1+v+w+x+y=12[/tex]
[tex]v+w+x+y=11[/tex]
no of solutions for this equation is
[tex]^{11+4-1}C_{4-1}=^{14}C_3[/tex]
if one solution is chosen at random Probability that u=1
[tex]=\frac{^{14}C_3}{^{16}C_4}[/tex]
[tex]=\frac{364}{1820}=0.2[/tex]
When a solution is chosen at random, the probability that u = 1 will be 0.2.
How to calculate the probability?
From the information given, it can be noted that the total number of solutions will be:
= (12+5+1)C(5-1)
= 16C4 = 1820
For, u = 1, the number of solutions will be:
= (11 + 4 - 1)C(4 - 1)
= 14C3
= 364
In conclusion, the probability will be:
= 364/1820
= 0.2
Learn more about probability on:
https://brainly.com/question/25870256
