# Robot and Probability

Robot drags parts from the conveyor to the box, and the number of parts it drags is randomly selected from $1$ to $k$. He drags the parts until there are $\ge n$ parts in the box. What is the expected value $\Epsilon \left[ X \right]$ of such actions?

Note by Ilya Pavlyuchenko
9 months, 3 weeks ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

First idea: find the number of system solutions for each natural $m \le n$, where $1 \le k_i \le k$ is number of the dragged parts in $i$ action: $\begin{cases} { k }_{ 1 }+{ k }_{ 2 }+\dots +{ k }_{ m }\ge n \\ { k }_{ 1 }+{ k }_{ 2 }+\dots +{ k }_{ m-1 } For $m = 1$ system is equivalent to inequality: $k_1 \ge n;$Then use the formula of expected value, where $N_{i}$ is solutions number of system above for $m = i$: $\Epsilon \left[ X \right] =\sum _{ i=1 }^{ n }{ { x }_{ i }{ p }_{ i } } =\sum _{ i=1 }^{ n }{ i\cdot \frac { { N }_{ i } }{ \sum _{ j=1 }^{ n }{ { N }_{ j } } } ; }$ But i don't know how to solve system of inequalities. Please, help to understand. And It's interesting to see your ideas!

- 9 months, 3 weeks ago

I think I would basically use your approach and use a computer to evaluate all the possibilities

- 9 months, 3 weeks ago

Are you asking for the expected value of the number of actions, or the expected number of parts in the box? (What is $X$?)

- 9 months, 2 weeks ago

I asking for the expected value of the number of actions. And interesting how to solve system of inequality above.

- 9 months ago

Instead of $\sum_{j=1}^n N_j$, we want the number of combinations of actions from $i$ actions. (We are doing: Actions we care about divided by possible actions, and summing over the number of possible actions).

Call this value $T_i$. So $E[X] = \sum_{i=1}^n i \times \frac{N_i}{T_i}$.

Now you must think of a possible way to get the value of $N_i$ and determine $T_i$.

- 9 months ago

Pretty interesting question. For me, math is just like games. I think I need to read more about this gold question and then I will solve the issue. Everyone has their own games, right?)

- 1 month, 1 week ago

AJAP1 is a complete transmembrane protein with 411 amino acid residues and its structure includes a separable N-terminal signal peptide (residues 1-43), extracellular domain (residues 44-282), transmembrane domain (residues 283-303), and intracellular cytoplasmic domains (residues 304-411).

https://www.creative-biogene.com/genesearch/AJAP1.html

- 8 months, 2 weeks ago

This isn't the right forum.

- 8 months, 1 week ago