Answer:
A)
Given that,
On each trial, the machine outputs a ball with one of the digits 0 through 9 on it.
To find the probability of getting 6.
In each trail the probability of getting 6 is,
[tex]p=\frac{1}{10}[/tex]Let X be the event of getting 6.
we get
[tex]P(X=6)=nC_r(p)^r(q)^{n-r}[/tex]we get,
[tex]q=1-p=1-\frac{1}{10}=\frac{9}{10}[/tex]where n=50 and r=8
n is the total number of trials
r is the sucess trail.
Substitute the values we get,
[tex]P(X=6)=50C_8(\frac{1}{10})^8(\frac{9}{10})^{50-8}[/tex][tex]=50C_8(\frac{1}{10})^8(\frac{9}{10})^{42}[/tex]Hence the required probability is,
[tex]=50C_8(\frac{1}{10})^8(\frac{9}{10})^{42}[/tex]On simplifying we get,
[tex]=0.06487\approx0.06[/tex]Answer is: 0.06