Link: https://www.gregmat.com/problems/quant?problem=a-serial-thief-gets-caught-once-on-average
For this problem, I am confused why it is assumed that the thief would get caught on the 10th attempt. That is not how probability works. It is not like he steals 9 times and thus it is “due” that he gets caught on the 10th. In fact, he has a 34% of not getting caught on the 10th attempt (0.9^10).
So simply dividing 181/9 is not enough.
The odds the theif gets away with stealing is ~0.53 on the 6th attempt (0.9^6), and after that the probability is less than 50%. Thus it should be better we consider only the 6th attempt and do 181/6 to get 30.1666 Euros, or 30 (Option E).
Why does the solution assume he literally gets caught 9 times out of 10, when the value is just an average?
I don’t see how your argument is related to the question at hand. As for your confusion with the thief getting caught on the 6th/10th attempt: we’re talking about the expected value (long-run average) and not just a single sample or a specific streak.
You can imagine a large number of simulations where the thief shoplifts thousands of times. Each attempt is an iid (independent and identically distributed) random variable. According to the Law of Large Numbers, the sample mean will converge to the expectation provided by the problem (1 out of 10). Just like if you were to randomly consider the outcome of 10-coin flips, you would likely not get exactly half heads and half tails. But if we shift our focus to a large number of independent random samples, then we can more confidently work with a 50% probability. It’s the same gist here. After all, he’s a “serial” thief.
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