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$A$ 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.

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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 $\frac{1-a}{2}$.

B) This yields to an analysis very similar to the analysis for your previous Problem 2, so I will leave it to you.

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  • $\begingroup$ @learner_avid: I expect that after the hint, you will not have difficulty with B. If there is, please send a message, But I will not be answering until tomorrow, it is very late here. $\endgroup$ – André Nicolas Oct 8 '15 at 7:13
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    $\begingroup$ @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: $P(n_P<n_Q)=P(n_P>n_Q)$. $\endgroup$ – drhab Oct 8 '15 at 7:30
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    $\begingroup$ Yes, that's what I am saying. $\endgroup$ – drhab Oct 8 '15 at 7:48
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    $\begingroup$ 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 $\frac{1-a}{2}$. $\endgroup$ – André Nicolas Oct 8 '15 at 14:39
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    $\begingroup$ 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 $(3/4)^n$. $\endgroup$ – André Nicolas Oct 8 '15 at 14:53

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