Question

The possibility that a salesman will lose his glasses while visiting a store is 1/5. Suppose a salesman visits two stores and loses his glasses. What is the probability that he lost his glasses in the first store?

Answer 1

20%

Answer 2

That's supposedly wrong. The answer should be 5/9

Answer 3

The problem can be reworded as follows:
'Given that the salesman lost his glasses when visiting one out of two stores, what is the probability that he lost the glasses in the first store he visited rather than the second store. The probability that he loses his glasses when visiting a store is 1/5.'

The events 'lost glasses in first store' and 'lost glasses in second store' are mutually exclusive (he can't have lost his glasses in both stores). Therefore:
P(lost in 1st store) + P(lost in 2nd store) = 1
His first opportunity to lose his glasses is in the first store he visited. Therefore P(lost in 1st store) will be greater than P(lost in 2nd store). Can you take it from there?

Answer 4

No I'm not sure I can take it from here. I understand that the salesman won't lose his glasses in the second store he he already lost them in the first, hence there should be a higher chance that he lost them at the first store. I'm not really sure how to calculate that difference however.

Answer 5

Let P = probability glasses lost in a store
Probability glasses lost in first store visited = P
Probability glasses lost in second store = Probability glasses not lost in first store visited * Probability glasses lost in a store = 4/5 * P
\[P+(\frac{4}{5})P=1\]
Solve for P to find the answer.

Answer 6

Thanks I eventually managed to work this out last night. But can I ask you, what is the difference between the above problem and the following:

"A man will go cycling 3 random mornings out of the week. When he goes cycling he has a 7/10 probability of having eggs for breakfast. When he does not go cycling, he has a 5/20 probability of having eggs for breakfast. Provided he cycled this morning, what it the probability that he had eggs for breakfast?"

Shouldn't this be: "P(cycling+eggs) \P(sum of all egg probabilities)"? That is 21/38? I don't understand why the 2 problems are different.

Answer 7

In the eggs for breakfast problem, if it is given that the man went cycling this morning I think that the probability he had eggs for breakfast is 7/10. However I am having trouble getting my head around this problem. I suggest that you post it as a question and invite @Zarkon to reply.