Real numbers $\{x_1,x_2,...\}$ are independently draw from the $[0,1]$. Then:

$a)$ $P(x_1+x_2<1\; |\; x_1<\frac17)=\frac{13}{17}$

$b)$ $P(x_1<\frac17 \;|\;x_1+x_2<1)=\frac{13}{49} $

$c)$ $P(x_1<\frac17 \;|\;x_1+x_2<1)=\frac{15}{49} $

$d)$ Probability that among numbers $\{x_1,x_2,...,x_8\}$ there are 6 numbers smaller than $\frac15$ is: $8\choose 6$*$(\frac15)^6*(1-\frac15)^2$

$e)$ $P(x_1+x_2<1)=\frac{2}{9}$

I have to check if these are true or not. I don't even know how to start this exercise. Any help will be much appreciated.

  • $\begingroup$ As stated either b) or c) must be false. For d) does 6 mean exactly 6 or 6 or more? $\endgroup$ – herb steinberg Jan 18 '18 at 17:23

Parts (a),(b),(c) and (e) are all quite similar, so I'll do (a) and hopefully you can complete the others. For part (d), @herb is right, it depends on whether it means exactly 6 or 6 or more. For more insight on (d), try to think about it as a binomial random variable.

By the definition of conditional probability:

$$P(x_1 + x_2 < 1 \, | \, x_1 < \frac{1}{7}) = \frac{P(x_1 + x_2 < 1 \text{ and } x_1 < 1/7)}{P(x_1 < 1/7)}.$$

The denominator is $1/7$, so we just need to solve for the numerator. Since $x_1$ and $x_2$ are independent and uniform, this is just the area of the region $$\{(x,y) \in [0,1]\times[0,1] : x + y < 1, x < 1/7\}.$$ We can use planar geometry, or just calculate it as an integral:

$$\int_0^{1/7} \int_0^{1 - x} dy\,dx = \int_0^{1/7} (1 - x)\,dx = \frac{13}{98}.$$

Dividing this by $1/7$ gives $$P(x_1 + x_2 < 1 \, | \, x_1 < \frac{1}{7}) = \frac{13}{14}\,.$$

  • 2
    $\begingroup$ +1 and definitely the answer. My post intended as a complementary approach. $\endgroup$ – Antoni Parellada Jan 18 '18 at 23:59
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    $\begingroup$ Thank you, for helping me with (a), I've already done (b),(c),(e) too. Regarding (d) it means exactly 6 not 6 or more. I'm currently working on it, but don't know how to do it yet. If you could write some more tips to (d), I will be grateful (even more). $\endgroup$ – MacAbra Jan 19 '18 at 10:04
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    $\begingroup$ Isn't it just: $8 \choose 6$ $*(\frac15)^6*(1-\frac15)^2$ $\endgroup$ – MacAbra Jan 19 '18 at 10:24
  • $\begingroup$ Yes exactly!${}{}{}{}{}{}$ $\endgroup$ – Marcus M Jan 20 '18 at 0:05

As an illustration of the planar geometry that Marcus M is referring to, and for the first exercise, here is the area to calculate for the numerator (in dashed purple lines):

enter image description here

We can calculate it as the area of the vertical rectangle under the purple dashed lines, and excluding the red square

$$\text{base} \times \text{height} =\frac{1}{7}\left(1- \frac{1}{7}\right)$$

plus the 1/2 of the red square:



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