# What is the probability that number of elements in $P$ is more than that in $Q?$

$A$ is a set containing $n$ elements,a subset $P$(may be void also) is selected at random from set $A$ and the set $A$ is then reconstructed by replacing the elements of $P.$A subset $Q$(may be void also)of $A$ is again chosen at random.

$(A)$What is the probability that number of elements in $P$ is more than that in $Q?$
$(B)$What is the probability that $Q$ is a subset of $P?$

I asked similar question and understood how to find the probability and on the basis of that i tried these questions but my answers are coming wrong.Please help me.

A) For the probability $P$ has more elements than $Q$, recall from the answer to the earlier problem about the probability $P$ and $Q$ are of equal size is $a$, where $a=\frac{\binom{2n}{n}}{2^{2n}}$.
So the probability $P$ and $Q$ are unequal is $1-a$, and therefore by symmetry the probability $P$ has more than $Q$ is $\frac{1-a}{2}$.
• @learner_avid Because $1-a=P(n_P>n_Q)+P(n_P<n_Q)=P(n_P>n_Q)+P(n_P>n_Q)=2P(n_P>n_Q)$ on base of symmetry: $P(n_P<n_Q)=P(n_P>n_Q)$. – drhab Oct 8 '15 at 7:30
• Abour the $\frac{1-a}{2}$. Imagine that player Paul wins if set $P$ is bigger than $Q$, player Quincey wins if $Q$ is bigger than $P$, and they tie if $P$ and $Q$ have the same number of elements. In an earlier problem we saw that the probability they tie is $\binom{2n}{n}/2^{2n}$. Call this number $a$. So the probability they don't tie $1-a$. But Paul and Quincey have equal chances of winning, so the probability Paul wins is $\frac{1-a}{2}$. – André Nicolas Oct 8 '15 at 14:39
• B: line up the elements of the set $A$, Together, Paul and Quincey each stop in front of an element $k$ of $A$. Each flips a fair coin. If Paul's lands heads, then $k$ gets chosen for $P$. If Quincey's lands heads, $k$ gets chosen for $Q$ (a number can be chosen for both, or neither, or just one). We want $Q$ to be a subset of $P$, so the only "bad" thing is $k$ chosen for $Q$ but not for $P$. This has probability $1/4$. So the probability that Paul and Quincey's decision on $k$ is "good" for $Q$ subset of $P$ is $3/4$. The probability they make $n$ good decisions in a row is $(3/4)^n$. – André Nicolas Oct 8 '15 at 14:53