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].

E[X] = 7/9

Step-by-step 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.