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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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We could find the slope with this formula
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NUMBER 20
Given:
(x₁,y₁) = (-2,3)
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Solution:
Input the points to the formula
m = (y₂ - y₁) / (x₂ - x₁)
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m = -7 / (7+2)
m = -7/9

The slope of the line is -7/9


NUMBER 21
Given:
(x₁,y₁) = (-6,-1)
(x₂,y₂) = (4,1)

Solution:
Input the points to the formula
m = (y₂ - y₁) / (x₂ - x₁)
m = (1-(-1)) / (4 -(-6))
m = (1+1) / (4+6)
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NUMBER 22
Given:
(x₁,y₁) = (-9,3)
(x₂,y₂) = (2,1)

Solution:
Input the points to the formula
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6 0
3 years ago
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit  10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

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40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is \overline A=\textrm{Missing a target at least once}

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, P(A)=0.95^{10}=0.5987

We know that the sum of probability of an event and its complement is 1.

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6 0
3 years ago
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I hope this helps you

6 0
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strojnjashka [21]

Hello from MrBillDoesMath!

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