Real numbers $\{x_1,x_2,...\}$ are independently draw from the $[0,1]$, then: 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. 
 A: 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 $$P(x_1 + x_2 < 1 \, | \, x_1 < \frac{1}{7}) = \frac{13}{14}\,.$$
A: 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):

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:
$$\frac{1}{2}(\text{side})^2=\frac{1}{2}\left(\frac{1}{7}\right)^2$$
