Answer:
0.057
Step-by-step explanation:
Given that:
Number of machines, [tex]n[/tex] = 10
Probability, that an individual machine will break down, [tex]p[/tex] = 0.1
Probability, that an individual machine will not break down, [tex]q=1-[/tex][tex]p[/tex]
To find:
The probability that on a given day, three machines will break down = ?
Solution:
First of all, let us have a look at the probability formula:
[tex]P(x=r) = _{n}C_{r}p^rq^{n-r}[/tex]
Here, r = 3
Putting the values:
[tex]P(x=3) = _{10}C_{3}p^3q^{10-3}\\\Rightarrow P(x=3) = _{10}C_{3}{0.1}^30.9^{7}\\\Rightarrow P(x=3) = 120\times{0.1}^30.9^{7}\\\Rightarrow P(x=3) = \bold{0.057}[/tex]
