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jeka94
3 years ago
10

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo

graphical errors on a page of second booklet is a Poisson random variable with mean 0.3. Suppose the number of errors on different pages of the same or different booklets is independent. There are 7 pages in the first booklet and 5 pages in the second one. Find the probability of more than 2 typographical errors in the two booklets in total.
Mathematics
1 answer:
muminat3 years ago
4 0

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: {\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

Where \lambda is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}

Substitute the respective values in the above formula.

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!}})

P(k > 2)= 1 - (0.44596)

P(k > 2)=0.55404

Hence, the required probability is 0.55404.

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