Help on Probability Problem

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.

Note by Jay B
1 year, 3 months ago

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Hi 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 an hour is the same event as not catching a fish in the first half hour and not catching a fish in the second half hour.
• Like you said, those two events (not first half, not second half) are independent due to uniform/independence conditions.
• Thus, P(A and B) = P(A)*P(B).
• By symmetry/uniformity, P(A) = P(B); i.e., there is no difference between the first half hour and second half hour.
• So, P(A and B) = P(A)$$^2$$.
• But no fish in an hour is 100%-64% = 36%. So P(A)$$^2$$ = 36%, so P(A) = 60%.

Staff - 1 year, 3 months ago

thanks!

- 1 year, 3 months ago