Let Pn be the set of all subsets of the set [n]={1,2,…,n}. If two distinct elements of P5 are chosen at random, the expected number of elements (of [n]) that they have in common can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?

136 wrong

To find the expected number of elements that two distinct elements of P5 have in common, we can use the principle of linearity of expectation.

Let's consider the j-th element of [5]={1,2,3,4,5}. The probability that this element is present in any given subset of P5 is 1/2, since there are 2^5=32 total subsets and each element has an equal probability of being included or excluded in any given subset.

Now, let's consider the number of elements in [5] that are common to two randomly chosen subsets. We can break this down into the following cases:

1. When j-th element is present in both subsets: The probability of this happening is (1/2)^2 = 1/4 because each subset independently includes or excludes the j-th element.

2. When j-th element is present in one subset but not the other: The probability of this happening is 2 * (1/2) * (1/2) = 1/2. The factor of 2 comes from the fact that the j-th element can be present in the first subset and absent in the second subset, or vice versa.

3. When j-th element is absent in both subsets: The probability of this happening is (1/2)^2 = 1/4 because each subset independently includes or excludes the j-th element.

Therefore, the expected number of elements that two distinct subsets of P5 have in common is given by the sum of the probabilities times the number of elements in each case:

E = (1/4) * 1 + (1/2) * 0 + (1/4) * 1
= 1/4 + 0 + 1/4
= 1/2

Since a and b are coprime positive integers, a+b = 1+2 = 3.

Therefore, the value of a+b is 3.