Answer:
The required probability is 0.55404.
Step-by-step explanation:
Consider the provided information.
The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.
Average error for 7 pages booklet and 5 pages booklet series is:
λ = 0.2×7 + 0.3×5 = 2.9
According to Poisson distribution: [tex]{\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}[/tex]
Where [tex]\lambda[/tex] is average number of events.
The probability of more than 2 typographical errors in the two booklets in total is:
[tex]P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}[/tex]
Substitute the respective values in the above formula.
[tex]P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})[/tex]
[tex]P(k > 2)= 1 - (0.44596)[/tex]
[tex]P(k > 2)=0.55404[/tex]
Hence, the required probability is 0.55404.
