A2. Reviewed statistics indicated that there is an average of five thefts per day from a large
supermarket operating from 08 00 to 19 00 six days a week. Find the probability that there
(a) are no thefts in a week. [2]
(b) will be ten thefts in any given day. [4]
(c) are four thefts by 10 00 in any given day. [4]
To find the probabilities in each scenario, we need to use the concept of probability distributions. In this case, we can use the Poisson distribution since we are dealing with the number of thefts occurring in a given time period.
The Poisson distribution is often used to model the number of events that occur in a fixed interval of time or space when the average number of events is known and the events occur independently of each other.
To solve each part of the question, we will use the following formula:
P(X = x) = (e^(-λ) * λ^x) / x!
Where:
- P(X = x) is the probability of x events occurring
- lambda (λ) is the average number of events
- e is the base of the natural logarithm (approximately 2.71828)
- x! represents the factorial of x
Now let's solve each part of the question:
(a) The average number of thefts per day is 5. Since there are 6 days in a week, the average number of thefts in a week would be 5 * 6 = 30.
Using the formula, we can calculate the probability of no thefts in a week:
P(X = 0) = (e^(-30) * 30^0) / 0!
Simplifying this expression, we get:
P(X = 0) = e^(-30) ≈ 9.36e-14
Therefore, the probability that there are no thefts in a week is approximately 9.36e-14.
(b) Now we want to find the probability of there being ten thefts in any given day. Since the average number of thefts per day is 5, we can directly use the Poisson distribution formula with λ = 5 and x = 10:
P(X = 10) = (e^(-5) * 5^10) / 10!
Evaluating this expression, we get:
P(X = 10) ≈ 0.018
Therefore, the probability that there will be ten thefts in any given day is approximately 0.018.
(c) Finally, we need to find the probability of there being four thefts by 10:00 in any given day. Here, we can adjust the average number of thefts using the time interval.
Assuming the average number of thefts per hour is 5/11 (since the supermarket operates for 11 hours), we can again use the Poisson distribution formula with λ = 5/11 and x = 4:
P(X = 4) = (e^(-5/11) * (5/11)^4) / 4!
Evaluating this expression, we get:
P(X = 4) ≈ 0.195
Therefore, the probability that there are four thefts by 10:00 in any given day is approximately 0.195.
Note: It is important to remember that the Poisson distribution assumes the events occur independently of each other.