Question

The 10 decimal digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are arranged in a uniformly random permutation. We denote by a the integer formed in base 10 by the first five positions in this permutation and by b the integer formed in base 10 by 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

a*b = odd only if both a and b are odd

for a to be odd : 5th digit in integer (last digit of a) = odd

for b to be odd : 10th digit in integer (last digit of b) = odd

P(both and b odd) = P(a odd)*P(b odd | a is odd)

= P(5th digit odd)*P(10th digit odd | 5th digit is odd)

= [ (no. of odd digits)/(total digits) ] * [ (no. of odd digits remaining)/(total no. of digits remaining) ]

= [5/10] * [4/9]

= 20/90

P(odd) = 0.2222

P(even) = 1 - P(odd) = 1 - 0.2222 = 0.7778

P(x=0) = P(odd) = 0.2222

P(x=1) = P(even) = 0.7778

E(x) = sum of [x.P(x)]

= 0*0.2222 + 1*0.7778

E(x) = 0.7778

P.S. (please upvote if you find the answer satisfactory)