Intro Independent Probability Question

AI Thread Summary
The probability of a student losing his library card on Thursday, given he lost it by that day, is calculated using the formula P(T|L) = P(T)/P(L). The process involves recognizing that each day is mutually exclusive for losing the card, and the probability of losing it on Thursday is derived from the probabilities of not losing it on the previous days. Regardless of the specific library visited, the probability of losing the card remains constant at X each day. When considering the likelihood of losing the card, Monday emerges as the day with the highest probability since it is the only day where the card cannot have been lost previously. Thus, the student should search for his card at the library he visited on Monday.
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Homework Statement



The probability of a student losing his library card is X each time he goes to the library. He goes to a different library 5 times a week, starting from Monday and ending on Friday.

1)At the end of the day on Thursday, he finally checks his wallet and realizes his library card is gone! What is the probability that the student lost his library card on Thursday and not any other day?

2)Which Library should he go to to search for his library card?

The Attempt at a Solution



Is this the correct process?

First see each day is mutually exclusive since once you lose the card that's it. You can't lose it on another day.

1)

P(T)=Probability of losing card on Thursday = (1-X)(1-X)(1-X)(X)

P(L)=Probability of him losing the card in those 4 days= 1-[(1-X)(1-X)(1-X)(1-X)]

P(T|L)=Probability of him losing the card on Thursday, given that he lost the card

= [(P(T)]/P[L) the answer were looking for

2) Doesn't matter which one he goes to, since the probability for each day is always X.

Any help is appreciated. It concerns me a bit I find introductory probability harder than Calc III.
 
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How does your answer for #2 look if X=0.9 ?
 
Think about what the probability is of losing it on a particular day. The day with the highest probability is the one he should visit.

Monday: X
Tuesday: (1-X)X
Wednesday: (1-X)2X
Thursday: (1-X)3X

If you work it out, you find the sum of these 4 probabilities is equal to 1 - (1-X)4, which is a nice sanity check.

X and any power of (1-X) are fractional. This makes the answer Monday, no matter what X is (unless it is 0). The reasoning is that you can only lose your card if you have not already lost it. So in this example, Monday is the only day of the four where the card is guaranteed not to be lost, making it the most likely day on which he lost his card. The other days are less likely since they are contingent on the boy not losing his card on the previous days.
 
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