1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ +1 and definitely the answer. My post intended as a complementary approach. $\endgroup$ – Antoni Parellada Jan 18 '18 at 23:59
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
3
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.