Question

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. 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.

Answer 1

The required probability is 0.55404.

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.