Find Probability of 3 Samples from Chi-Square Distribution Exceeding 7.779

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Homework Statement



If 15 Observations are taken independently from a chi-square distribution with 4 degrees of freedom, find the probability that at most 3 of the 15 sample items exceed 7.779

Homework Equations


The Attempt at a Solution



This problem should be quite simple. I find that in the back of my text that the chi square distribution of 7.779 is .9... but the answer is .9444... I am at a loss at why it is not exactly .9 and where to go from here. I would appreciate any help. Thanks.
 
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Have you taken into consideration that the problem asks for "at most 3"?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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