Sets + Probability = aweSUM !

Probability Level 4

$A$ is a set consisting of 10 elements. Subsets $P$ and $Q$ are selected at random (the subsets can be the same). Find the probability such that $Q$ has just 1 more element than $P$ .

Note:
The null set $\emptyset$ is also a valid subset.

The answer is 0.1601.

1 solution

Avineil Jain
May 7, 2014

Let r elements be selected from A for subset P. Number of ways of selecting P is-

$\dbinom{n}{r}$

Since the set A is complete again, we can select Q in-

$\dbinom{n}{r+1}$ ways

Total number of ways -

$\displaystyle\sum_{i=0}^k \dbinom{n}{r} \dbinom{n}{r+1}$ where $k= n-1$

Using properties of binomial, we can easily find the sum. It comes out to be $\dbinom{2n}{n+1}$

Since total number of ways of selecting P and Q is $4^{n}$ , the probability is-

$P(E) = \dfrac{\dbinom{2n}{n+1}}{4^{n}}$

Substitute n=10 to get the answer!

There is no need to mention that "the elements of A are replaced", because they are not "removed". I've updated the wording of the problem. Please check that it is accurate.

Calvin Lin Staff - 7 years, 1 month ago

OK Sir. I just wanted to be more specific!

Avineil Jain - 7 years, 1 month ago

@Calvin Lin @Avineil Jain Can't the number of ways of selecting $P$ and $Q$ be $\binom{2^{10}}{2}$ , as there are a total of $2^{10}$ subsets, instead of $\left(\binom{10}{0}+\binom{10}{1}+ \cdots+\binom{10}{10}\right)^2$ ?

The problem should mention where to select $P$ and $Q$ from, that is, from the power set of $A$ or from the set $A$ itself.

That is, it should mention whether $P$ can be equal to $Q$ or not.

Pratik Shastri - 6 years, 8 months ago

I've updated the phrasing to indicate that we could have $P=Q$ .

P and Q are subsets of A, which are also elements of the power set of A.

Calvin Lin Staff - 6 years, 8 months ago

I meant that if we select $P$ and $Q$ from the power set of $A$ , they will be different and if we select them from $A$ , then they can be equal.

Pratik Shastri - 6 years, 8 months ago

I thought Q should have all the elements of P plus one additional element, which gives another answer

Lina Sharifi Moghaddam - 6 years, 6 months ago

Done Exactly the Same Way! the answer comes out to be: $\large\dfrac{20995}{131072}$

Kunal Gupta - 6 years, 2 months ago

Sets + Probability = aweSUM !

The answer is 0.1601.

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