# 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
2 years, 9 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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 - 2 years, 9 months ago

thanks!

- 2 years, 9 months ago