Calculating Probability for Two Events: Longer Than 24.5 Minutes

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The discussion revolves around calculating the probability of two events being longer than 24.5 minutes. The initial assumption was that the probability would be calculated using the values 8/55 and 7/54, but the correct calculation involves 11/55 and 10/54. The confusion stems from determining how many ways to choose 2 events from the group exceeding 24.5 minutes. The participants clarify the combinatorial aspects of the problem, emphasizing the importance of correctly identifying the groups involved. The conversation concludes with an acknowledgment of the misunderstanding.
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As the question says the probability that both of them were longer than 24.5 minutes i'd of thought it would be in the 24-29 row, so that means 8/55 * 7/54, but the answer is 11/55 * 10/54, where did they get 11 from?
 
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How many ways are to choose 2 from the over 24.5 minute group? How many ways are there to choose 2 overall?
 
LCKurtz said:
How many ways are to choose 2 from the over 24.5 minute group? How many ways are there to choose 2 overall?

oh god I am stupid, thank you.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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