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My attempt: Have $5$ random variables uniform on the interval $[0, 1]$, with cdf $F(x) = x$.

The middle half of the interval is $[\frac{1}{4}, \frac{3}{4}]$. The probability that one of the random variables lies in this interval is

$$ p = F\Big(\frac{3}{4}\Big) - F\Big(\frac{1}{4}\Big) = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} .$$

The probability that all five random variables lie in this interval is then

$$ p^5 = \Big(\frac{1}{2}\Big)^5 = \frac{1}{32} .$$

There was no solution for this exercise in my textbook and I couldn't find the answer anywhere online. Hoping somebody can verify if this is the correct answer.

  • 1
    To me it seems correct assuming independence of the trials. – 2017-02-02
  • 2
    looks good to me – 2017-02-02
  • 0
    P(X=5) in a Binomial(n=5,p=1/2)... – 2017-02-02

2 Answers 2

5

Your solution looks entirely correct to me. Strictly speaking, the phrase

$5$ numbers chosen randomly in the interval $[0,1]$

is a little vague, because it doesn't say if the picks are independent, but you can safely assume that, and then the probability of each event is $\frac12$, so the probability of them all is $\frac{1}{32}$. Well done!

1

Don't even need the ' F(x)' stuff since the interval u want them to be in has length 1/2 and the other two intervals [0,1/4) and (3/4,1] combined are length 1/2 so the probability will be (1/2)^n for n bring the numbers of picked .this case is n=5 .