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Let $Y \sim \mathcal{U}(0,4)$. If 20 independent random samples are extracted, what is the probability that in at least 5 of them $Y > 2$?

My attempt was: the required probability should be given by $1 - P(Y \leq 2)^5$, which would result in $\boxed{\boxed{P^\star = 1-0,5^5 = 0.96875}}$. However, when simulating the problem using R, I find that the probability is around $0.993$ or $0.994$.

Could anyone help me figure out where I am going wrong?

Thanks!

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    $\begingroup$ The simulation is correct and your calculation is not. If you have learned about the binomial distribution, try thinking about it that way. Each one has a 50% chance of being greater than 2. It's like flipping 20 coins and finding the probability that you see at least 5 heads out of the 20. $\endgroup$
    – John L
    Mar 19, 2021 at 17:43
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    $\begingroup$ Yes, indeed. I was able to see that. The correct calculation should be $1 - 0,5^{20} \cdot \left[ \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \binom{20}{3} + \binom{20}{4} \right] \approx 0,99409$, is that right? Thanks! $\endgroup$ Mar 19, 2021 at 17:47
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    $\begingroup$ Yes that is correct. $\endgroup$
    – John L
    Mar 19, 2021 at 18:17

1 Answer 1

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As pointed out in the comments, the correct answer is given by: $$ 1 - 0,5^{20} \cdot \left[ \binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \binom{20}{3} + \binom{20}{4} \right] \approx 0,99409 $$

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