Question

10 digits integers uniformly random permutation. five positions the last five positions in this permutation (either a or b may begin with 0 which in such a case is ignored). For example, if the random permutation is 8621705394 then a = 86217 and b = 5394. Consider the probability space whose outcomes are these random permutations and a random variable X defined on this probability space such X = 1 when the product ab is even and X = 0 when that product is odd. Calculate E[X].

Answer 1

E[X] = 7/9

Explanation:

We may assume that we obtain a success if the product ab is even and a failure if it is odd (a success if X =1, and a failure if X = 0). This means that X is a bernoulli random variable. The mean of X is equal to the probability of success in this experiment.

The product of 2 numbers is even if at least one of them is even. In order for a nmber to be odd or even we only need to care about its last digit: we can ignore the other 4. Therefore, we can only look at the last digit of each number and check whether there are even digits or odd ones.

We will compute P(X=0) because it is easier than directly calculating P(X=1), and also because P(X=1) = 1-P(X=0) (becuase it is the complementary event). In order for X to be 0 we need ab to be odd, therefore, we need the last digit of both a and b to be odd numbers.

The probability of the last digit of a to be odd is 5/10 = 1/2 (because there are 10 digits and 5 of them are odd: 1,3,5,7 and 9). Once we know that a is odd, we can discard the last digit of a, thus we have 4 odd digits left from a total of 9 digits. That means that the probability of b to also be odd is 4/9.

Multiplying both probabilities, we obtain that the probability that both a and b (and, as a result, ab) is odd is 1/2*4/9 = 2/9. This means that P(X=0) = 2/9, and P(X=1) = 1-2/9 = 7/9.

As a result, and because X is a bernoulli random variable, we conclude that E[X] = 7/9.