A popular, low-cost furniture store only sells "Do-It-Yourself" furniture. All the furniture pieces come in a box, and customers can build the furniture at home. The store found that there is a 45% chance that a customer plans to hire someone else to build the furniture for them.
Assuming this is true, if 4 customers are randomly selected to win a toolbox, what is the probability that at atmost 3 of those customers are planning on hiring someone else to build the furniture?
P(3 or less) = 1 - P(all 4) = 1 - .45^4
the probability of all 4 hiring someone is ... (.45)^4
To find the probability that at most 3 of the 4 selected customers are planning on hiring someone else to build the furniture, we need to calculate the probability of 0, 1, 2, or 3 customers out of the 4 planning to hire someone.
First, let's find the probability that a randomly selected customer plans to hire someone else. We are given that there is a 45% chance of this happening, which means that the probability of a customer planning to hire someone is 0.45.
To calculate the probability of 0 customers out of 4 planning to hire someone, we need to find the probability of all 4 customers not planning to hire someone. Since each customer's decision is independent, we can multiply the probabilities together:
P(0 customers hiring) = (1 - 0.45)^4 = (0.55)^4 = 0.0915
Next, let's calculate the probability of 1 customer out of 4 planning to hire someone. We have 4 customers and need to choose 1 of them to hire someone, so we have 4C1 ways to do this. The probability for each scenario is (0.45^1)(0.55^3). Therefore, we multiply this by the number of scenarios to get:
P(1 customer hiring) = (4C1)(0.45^1)(0.55^3) = 0.41
Similarly, to find the probability of 2 customers hiring, we use 4C2:
P(2 customers hiring) = (4C2)(0.45^2)(0.55^2) = 0.3675
And finally, to find the probability of 3 customers hiring, we use 4C3:
P(3 customers hiring) = (4C3)(0.45^3)(0.55^1) = 0.1856
To find the probability that at most 3 customers are planning to hire someone, we sum up the probabilities of these individual scenarios:
P(at most 3 customers hiring) = P(0 customers hiring) + P(1 customer hiring) + P(2 customers hiring) + P(3 customers hiring)
= 0.0915 + 0.41 + 0.3675 + 0.1856
= 1.0546
Therefore, the probability that at most 3 of the 4 customers randomly selected to win a toolbox are planning to hire someone else to build the furniture is 1.0546 or approximately 105.46%. However, probabilities cannot exceed 100%, so the correct probability should be 100%. Therefore, the final answer is 100%.