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$p(x)=3x^2$ with $0 < x < 1$: 0 elsewhere.

$p(x)$ is the probability density function of the variables.

there are 3 independent variables $X_1$,$X_2$,$X_3$ with the distribution listed above

What is the probability exactly 2 are greater than $1/2$?


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If $p(x)$ is supposed to be the probability density function of each of the three random variables then you should probably say so. – Henry Sep 23 '11 at 22:05

Hint 0: for a continuous random variable $X$ with a probability density function $p(x)$, you have $\Pr(X \le k) = \int_{x=-\infty}^{k} p(x) \; \text{d}x$

Hint 1: Work out $\Pr\left(X_1 \le \frac{1}{2}\right)$ and $\Pr\left(X_1 \gt \frac{1}{2}\right)$; the same will apply to $X_2$ and $X_3$ as they have identical distributions

Hint 2: As they are independent and identical, treat the question as finding the probability of getting exactly two success from three in a binomial distribution

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Yes the 3 choose 2 is fine. That is not the tricky part. I am just having a difficult time thinking of the proper result for the pdf aspect of the question. p^x*((1-p)^(n-x)) * nCr i believe will be the second part. its just the p(x) aspect that i cannot recall. by the way this is not homework. just a brain teaser for myself. – jason m Sep 23 '11 at 22:05
@jason: Try my added hint 0 – Henry Sep 23 '11 at 22:11
@AndréNicolas Isn't the CDF $x^3$ for $0 \leq x < 1$ giving $P\{X_i > 1/2\} = 7/8$? – Dilip Sarwate Sep 23 '11 at 23:52
@Dilip Sarwate: Yes, thank you, I can't integrate. – André Nicolas Sep 24 '11 at 0:16

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