Condition for two events to be independent

Question:

An experiment has 10 equally likely outcomes. Let $A$ and $B$ be two non-empty events of that experiment. If $A$ consists of $4$ outcomes, find the number that $B$ must have in order for $A$ and $B$ to be independent.

I really don't understand what the question is asking. For two events to be independent, the probability of one of the events occurring must be independent of the other event occurring. So shouldn't the answer simply be $6$?

• Recall the definition of independence of events. – Clement C. May 13 '16 at 13:35
• I would say "at most 6". – user247327 May 13 '16 at 13:35
• @user247327 See, that's incorrect as well. That's why I got even more confused.... – Gummy bears May 13 '16 at 13:36
• @Gummybears The probability of $A$ is $\Pr[A]=[\dots]$. If $B$ contains $n$ outcomes, with $k\leq \min(4,n)$ become common with $A$, then $\Pr[A]=[\dots]$ and $\Pr[A\cap B]=[\dots]$. So what are the settings of $n,k$ that make independence (as per the definition linked above) possible? – Clement C. May 13 '16 at 13:37
• @Gummybears Read kccu's answer or my comment(s). You need $0\leq k \leq \min(n,4)\leq 10$ satisfying $\frac{4}{10}\cdot\frac{n}{10} = \frac{k}{10}$. – Clement C. May 13 '16 at 13:42

It sounds like you're confusing independent with mutually exclusive. Events $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$. Events $A$ and $B$ are mutually exclusive if $A \cap B = \varnothing$, i.e., whenever one of $A$ or $B$ occurs, the other does not.
This problem asks you to determine what the number of outcomes in $B$ must be in order to have $P(A \cap B)=P(A)P(B)$. Since the experiment has $10$ equally likely outcomes,
The equation $P(A\cap B)=P(A)P(B)$ then becomes $\frac{|A\cap B|}{10}=\frac{4}{10}\cdot \frac{|B|}{10}$, or $5\cdot |A\cap B|=2 \cdot |B|$. What are the possible values of $|B|$ that could satisfy this equation?
• This gives $2n = 5k$, writing $k=\lvert A\cap B\rvert$ and $n=\lvert B\rvert$. Since you also have $k \leq \min(n,4)$ and $n\leq 10$, the fact that $2n=5k$ leaves few options (in particular, $k$ has to be even, so $k\in \{0,2,4\}$). Check all 3 cases to see what happens. – Clement C. May 13 '16 at 13:44
• @Gummybears Yes, there are only a few cases to check, and as Clement C. points out you can rule some out pretty easily. Perhaps it's useful to note that $|B|$ must be divisible by $5$, which leaves just three possibilities to check. – kccu May 13 '16 at 13:46