# Probability of a random subset of Z

I'm stuck in this question, could someone give me a hand? I'll post what I've done so far.

Question 9: Let $$A=(1,2,3,4)$$ and $$Z=(1,2,3,4,5,6,7,8,9,10)$$, if a subset B of Z is selected by chance calculate the probability of:

a) $$P(B⊂A)$$ B is a proper subset of A

$$(3/10)*(2/10)*(1/10) =3/500$$ of chance

b)$$P(A∩B=Ø)$$ A intersection B =empty set

$$(2/10)*(3/10)*(4/10)*(5/10)*(6/10)=9/1250$$

Is it correct?

• Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. – 5xum Apr 23 at 13:45
• Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. – 5xum Apr 23 at 13:45
• Ok, thanks, I'll edit the qustion – AngryDog Apr 23 at 13:58
• Welcome to MSE. Please edit and use MathJax to properly format numbers and math expressions. – Lee David Chung Lin Apr 23 at 14:11
• How (accoring to which distributin) is the random subset selected? Uniformly over all subsets? – Hagen von Eitzen Apr 23 at 14:44

A better approach for a is to ask how many subsets $$Z$$ has and how many proper subsets of $$A$$ there are. Similarly for b you want a subset of $$A^c$$. How many of those are there?
• No, there are $1024$ subsets of $Z$. The empty set should count, as should all of $Z$, as the problem does not exclude them. Yes, there are $15$ proper subsets of $A$. $B$ is randomly selected from the $1024$ subsets, so the chance is $15/1024$ – Ross Millikan Apr 23 at 14:17
• No, for $A \cap B = \emptyset$ you need $B$ to not include any of $1,2,3,4$, so it need to be a subset of $\{5,6,7,8,9,10\}$ – Ross Millikan Apr 23 at 14:38
• So should I calculate the number of subsets of ${5,6,7,8,9,10}=2^6-1=63$ or should I multiply the chance of any element int the set like 2/10*3/10 like I did in the original question? – AngryDog Apr 23 at 14:45