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

Question

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.

Answer 1

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

Answer 2

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:

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