SOLUTION: Records show that the average number of bad checks that a bank receives in four days is 1 1.)What is the probability that 1 bad check will be received on any given day 2.)What is

Question 1169637: Records show that the average number of bad checks that a bank receives in four days is 1
1.)What is the probability that 1 bad check will be received on any given day
2.)What is the probability that bad checks will be received for any given week
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**Understanding the Problem**
We are given that the average number of bad checks received in four days is 1. We need to find the probability of receiving 1 bad check in a single day and the probability of receiving bad checks in a given week.
**Assumptions**
* We can model the number of bad checks received using a Poisson distribution.
* The rate of bad checks is constant.
* The number of bad checks received on different days are independent.
**Calculations**
1. **Average Bad Checks per Day:**
* The average number of bad checks in four days is 1.
* Therefore, the average number of bad checks per day (λ) is 1 / 4 = 0.25.
2. **Probability of 1 Bad Check on a Given Day:**
* We use the Poisson probability formula:
* P(X = k) = (e^(-λ) * λ^k) / k!
* Where:
* X is the number of bad checks.
* k is the number of bad checks we want to find the probability for.
* λ is the average number of bad checks per day (0.25).
* e is Euler's number (approximately 2.71828).
* We want to find P(X = 1):
* P(X = 1) = (e^(-0.25) * 0.25^1) / 1!
* P(X = 1) = (e^(-0.25) * 0.25) / 1
* P(X = 1) ≈ (0.7788 * 0.25)
* P(X = 1) ≈ 0.1947
3. **Probability of Bad Checks in a Given Week:**
* A week has 7 days.
* The average number of bad checks in a week is:
* λ_week = 0.25 checks/day * 7 days/week = 1.75 checks/week.
* The problem asks the probability of bad checks being received, this is vague. I will calculate the probability of recieving at least one bad check in a week. That is 1 - P(X=0)
* P(X = 0) = (e^(-1.75) * 1.75^0) / 0!
* P(X = 0) = e^(-1.75)
* P(X = 0) ≈ 0.1738
* P(X >= 1) = 1 - P(X=0) = 1 - 0.1738 = 0.8262
**Final Answers**
1. The probability that 1 bad check will be received on any given day is approximately 0.1947.
2. The probability that at least one bad check will be received for any given week is approximately 0.8262.

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