I got this problem on a course quiz:

In a certain lake, the probability of catching a fish is uniform and independent across time. If the probability that you catch at least one fish in an hour is 64%, what is the probability that you catch at least one fish in a half-hour?

I got it wrong and the solution is the following:

The probability that you don't catch a fish in one hour is 36%. **To not catch a fish in an hour, you have to not catch a fish in the first half-hour and in the second half-hour, so the probability of not catching a fish in a half-hour is \(\sqrt{36\%} = 60\%.\)** Thus, the probability of catching a fish in a half-hour is \(40\%.\)

I don't understand the stuff in bold print. What I think it means tough, is that not catching a fish in the fist half-hour and not catching a fish in the second half-hour are independent events because I think that this is what we mean by the probability of catching a fish is uniform; As another example, not catching a fish in 1/4 of an hour is independent from not catching a fish in 3/4 of an hour. Am I right with this? Moreover, if we call these events on the previous sentence A and B, respectively, then P(A and B) = P(not catch a fish in first 1/4 and not catch a fish in third quarter) = P(A)P(B) = 0.36. I need some verification. This is the best explanation I can come up with.

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TopNewestHi Jay --

In the future, if you believe a problem is flawed, you can report it directly on the problem. See here for details.

What the sentence in bold means is that:

not catching a fish in the first half hour and not catching a fish in the second half hour.Log in to reply

– Jay B · 1 month, 3 weeks ago

thanks!Log in to reply