The average number of advertisement per page appearing in a newspaper is 3. what is the probability that in a particular page zero number of advertisement are there?

Question

Answer

Probability of Zero Advertisements on a Newspaper Page

Given Information:

Average number of advertisements per page = 3

Solution:

Since the average number of advertisements per page is given as 3, we can assume that the distribution of the number of advertisements on a page follows a Poisson distribution with parameter λ = 3.

Let X be the random variable representing the number of advertisements on a page.

The probability of having zero advertisements on a page can be calculated as:

$$P(X=0) = \frac{e^{-\lambda}\lambda^0}{0!} = e^{-\lambda} = e^{-3} \approx 0.0498$$

Explanation:

The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space when these events occur independently and at a constant rate. In this case, the fixed interval of space is a newspaper page, and the events are the advertisements that appear on the page.

The Poisson distribution has one parameter, λ, which represents the average rate of events. In this case, λ = 3 since the average number of advertisements per page is 3.

The probability of having zero advertisements on a page is given by the Poisson probability mass function, which is:

$$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$$

Where k is the number of events (advertisements) and e is the mathematical constant approximately equal to 2.71828.

When k=0, the formula simplifies to:

$$P(X=0) = \frac{e^{-\lambda}\lambda^0}{0!} = e^{-\lambda}$$

Substituting λ = 3 into the formula, we get:

$$P(X=0) = e^{-3} \approx 0.0498$$

Therefore, the probability of having zero advertisements on a page is approximately 0.0498 or 4.98%.

Answer

E-3

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