The probability a car salesman sells a car to a customer is 0.05
Assuming the salesmen sees 14 customers in a week, what is the probability he sells less than 3 cars?
n = no sale , s = sale ... less than 3 is 0, 1, or 2
this is binomial ... no sale or sale ... p(n) = .95 ... p(s) = .05
(n + s)^14 = n^14 + 14 n^13 s + 91 n^12 s^2 + ... + 14 n s^13 + s^14
sum the 1st three terms
... .95^14 + (14 * .95^13 * .05) + (91 * .95^12 * .05^2)
To find the probability that the car salesman sells less than 3 cars, we can use the binomial probability formula. The formula for binomial probability is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where:
- P(X = k) is the probability of selling exactly k cars.
- C(n, k) is the number of combinations of n items taken k at a time (also known as the binomial coefficient).
- p is the probability of selling a car to a customer.
- n is the total number of customers seen by the salesman.
In this case, we want to find the probability of selling less than 3 cars, which means we need to calculate:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
Let's plug in the values into the formula:
P(X = 0) = C(14, 0) * 0.05^0 * (1 - 0.05)^(14 - 0)
Using the fact that C(n, 0) = 1, the calculation simplifies to:
P(X = 0) = 1 * 1 * (1 - 0.05)^14
We can use the same formula to calculate P(X = 1) and P(X = 2) by substituting the respective values of k.
Once we have obtained the probabilities for each scenario, we can sum them up to find P(X < 3) and get our final answer.