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 2 cars?
prob(sell car) = .05
prob(not sell car) = .95
Prob(less than 2 cars)
= prob(sell no car) + prob(sell 1 car)
= C(14,0) (.05)^0 (.95)^14 + C(14,1) (.05)^1 (.95)^13
= ...
let me know what you get
0.153
To find the probability that the car salesman sells less than 2 cars, we can use the binomial distribution formula.
The formula for the probability mass function (PMF) of the binomial distribution is:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X=k) is the probability of getting exactly k successes
- C(n, k) is the number of combinations of n items taken k at a time
- p is the probability of success
- n is the number of trials
In this case, the number of trials (n) is 14, as the car salesman sees 14 customers in a week. The probability of success (p) is 0.05, as the probability the salesman sells a car to a customer is 0.05. We want to find the probability of selling less than 2 cars, so k can be 0 or 1.
Let's calculate the probabilities for k=0 and k=1:
For k=0:
P(X=0) = C(14, 0) * 0.05^0 * (1-0.05)^(14-0)
Using the combination formula C(14, 0) = 1, this simplifies to:
P(X=0) = 1 * 1 * 0.95^14 = 0.4877
For k=1:
P(X=1) = C(14, 1) * 0.05^1 * (1-0.05)^(14-1)
Using the combination formula C(14, 1) = 14, this simplifies to:
P(X=1) = 14 * 0.05 * 0.95^13 = 0.3770
To find the probability of selling less than 2 cars, we can add the probabilities for k=0 and k=1:
P(X<2) = P(X=0) + P(X=1)
P(X<2) = 0.4877 + 0.3770
P(X<2) = 0.8647
Therefore, the probability that the car salesman sells less than 2 cars is approximately 0.8647.
To find the probability that the car salesman sells less than 2 cars in a week, we need to calculate the probability of selling 0 cars and 1 car, and then add those probabilities together.
Step 1: Calculate the probability of selling 0 cars.
The probability of selling 0 cars to a single customer is given as 0.05. Since the salesman sees 14 customers in a week, we can use the binomial probability formula to calculate the probability of selling 0 cars:
P(X = 0) = (n C x) * (p^x) * ((1-p)^(n-x))
Where:
n = number of trials (customers seen in a week) = 14
x = number of successes (cars sold) = 0
p = probability of success (selling a car to a customer) = 0.05
Using this formula:
P(X = 0) = (14 C 0) * (0.05^0) * ((1-0.05)^(14-0))
P(X = 0) = 1 * 1 * 0.95^14
P(X = 0) ≈ 0.4877
Step 2: Calculate the probability of selling 1 car.
Similar to the previous step, we use the binomial probability formula:
P(X = 1) = (n C x) * (p^x) * ((1-p)^(n-x))
Where:
n = 14 (number of trials)
x = 1 (number of successes)
p = 0.05 (probability of success)
Using this formula:
P(X = 1) = (14 C 1) * (0.05^1) * ((1-0.05)^(14-1))
P(X = 1) = 14 * 0.05 * 0.95^13
P(X = 1) ≈ 0.3946
Step 3: Add the probabilities.
To find the probability of selling less than 2 cars, we add the probabilities of selling 0 cars and 1 car together.
P(X < 2) = P(X = 0) + P(X = 1)
P(X < 2) ≈ 0.4877 + 0.3946
P(X < 2) ≈ 0.8823
Therefore, the probability that the car salesman sells less than 2 cars in a week is approximately 0.8823.