Final answer:
The odds that the salesman makes a sale at the first stop, the second stop, or both are calculated using the principle of inclusion-exclusion for probabilities. Given the probabilities of 36% for the first stop, 36% for the second stop, and 8% for both stops, the combined probability is 64%.
Step-by-step explanation:
The subject of this question is Mathematics, specifically dealing with the concept of probability. The question asks for the odds that a salesman will make a sale at his first stop, second stop, and both stops, given the probabilities for each scenario. To find the odds that the salesman makes a sale at either the first stop, the second stop, or both, we need to use the principle of inclusion-exclusion for probabilities. The formula to calculate this is:
P(A or B) = P(A) + P(B) - P(A and B)
Where P(A) is the probability of a sale at the first stop, P(B) is the probability of a sale at the second stop, and P(A and B) is the probability of a sale at both stops. Given P(A) = 36%, P(B) = 36%, and P(A and B) = 8%, the calculation would be:
P(A or B) = 0.36 + 0.36 - 0.08 = 0.64 or 64%
Therefore, the odds that the salesman makes a sale at the first stop, the second stop, or both are 64%.