The probability of at most three pages with errors is 0.4060021.
Given to us:
X follows a binomial distribution with,
Number of pages in the novel n = 400
Probability of error p = 0.005
Probability of page without error q = 01 - 0.005 = 0.995
The Probability Mass Function(PMF) for X is:
[tex]P(x)=\dfrac{n!}{(n-x)!x!}p^xq^{(n-x)\ =\ (\frac{n}{x} )p^xq^{(n-x)}[/tex]
a.) One of the 400-pages novels will contain exactly 1 page with errors
[tex]\begin{aligned}P(x=1)&=\ (\dfrac{n}{x} )p^xq^{(n-x)}\\\\&=(\dfrac{400}{1} )(0.005)^1(0.995)^{(400-1)}\\\\&=400\times 0.005 \times 0.995^{399}\\\\&= 0.2706694\end{aligned}[/tex]
b.) At most three pages are with errors
[tex]P(x\leq 3)= P(x=0)+P(x=1)P(x=2)+P(x=3)[/tex]
[tex]\begin{aligned}P(x=0)&=\ (\dfrac{n}{x} )p^xq^{(n-x)}\\\\&=(\dfrac{400}{0} )(0.005)^0(0.995)^{(400-0)}\\\\&=1\times 1 \times 0.995^{400}\\\\&= 0.13465\end{aligned}[/tex]
[tex]\begin{aligned}P(x=2)&=\ (\dfrac{n}{x} )p^xq^{(n-x)}\\\\&=(\dfrac{400}{2} )(0.005)^2(0.995)^{(400-2)}\\\\&=200\times 0.000025 \times 0.995^{398}\\\\&= 0.00068\end{aligned}[/tex]
[tex]\begin{aligned}P(x=3)&=\ (\dfrac{n}{x} )p^xq^{(n-x)}\\\\&=(\dfrac{400}{3} )(0.005)^3(0.995)^{(400-3)}\\\\&= 0.00000227\end{aligned}[/tex]
Therefore, the probability of at most three pages with errors is
[tex]\begin{aligned}P(x\leq 3)&= P(x=0)+P(x=1)P(x=2)+P(x=3)\\&=0.13465+0.2706694+0.00068+0.0000027\\&=0.4060021\end{aligned}[/tex]
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