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Mathematics

Probability

Poisson Distribution

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Assessing the probability that a bookstore sells 2 rare books in a week when it typically sells 4 rare books a week.

Compute $P(X=2) = \frac{e^{-4} \cdot 4^2}{2!}$ which gives $P(X=2) = \frac{16e^{-4}}{2}$ .

Calculating the likelihood of 0 accidents at a road crossing in a month when the average is 3 accidents per month.

The probability is $P(X=0) = \frac{e^{-3} \cdot 3^0}{0!}$ , resulting in $P(X=0) = e^{-3}$ .

The probability of a call center receiving 20 calls in an hour when they usually get 15 calls per hour.

Use the equation $P(X=20) = \frac{e^{-15} \cdot 15^{20}}{20!}$ to find the probability.

The likelihood of a rare flower blooming 8 times in a season when it usually blooms 3 times.

$P(X=8) = \frac{e^{-3} \cdot 3^8}{8!}$ computes the desired probability.

Determining the probability of observing 12 cars passing a checkpoint within an hour when the average is 10 cars per hour.

Using the formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ , where $\lambda = 10$ and $k=12$ , the probability is $P(X=12) = \frac{e^{-10} 10^{12}}{12!}$ .

Probability of receiving 5 emails in the next hour when on average 2 emails are received per hour.

Calculation: $P(X=5) = \frac{e^{-2} \cdot 2^5}{5!}$ , which simplifies to $P(X=5) = \frac{32e^{-2}}{120}$ .

Evaluating the probability of observing 10 particles emitted from a radioactive source in a minute if on average 6 particles are emitted per minute.

This is calculated by $P(X=10) = \frac{e^{-6} \cdot 6^{10}}{10!}$ .

Finding the probability of having exactly 3 system failures in a day when the average is 1 failure per day.

Use $P(X=3) = \frac{e^{-1} 1^3}{3!}$ , resulting in $P(X=3) = \frac{e^{-1}}{6}$ .

What is the probability that a 24/7 convenience store will have exactly 1 customer in 15 minutes during nighttime where the average is 0.5 customers per 15 minutes?

Apply $P(X=1) = \frac{e^{-0.5} \cdot 0.5^1}{1!}$ , leading to $P(X=1) = \frac{0.5e^{-0.5}}{1}$ .

Figuring out the chance of no alarms in a factory over a week when historically there are 2 alarms per week.

The probability is $P(X=0) = \frac{e^{-2} \cdot 2^0}{0!}$ , so $P(X=0) = e^{-2}$ .

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