Two digits are chosen at randomfrom a table of random numbers containing digits 0,1,2....9. Find the probability that
(a) the sum of the two digits is greater than 9 given that the first number is 3
(b) the second number is 2 given that the sum of the two munber is greater than 7
(c) the first number is 4 given that the difference between the two numbers is 4.
this is whati said:
let A denote " the sum of two digits chosen is greater than nine"
let B denote "the first digit chosen is 3"
P(B) = 9/10
but i dont know the possibility space for A
i know that P(A given B) = P(A n B)/P(B)
so can u please give me a push in the rite direction. and a general short explaination please.
Thank you so much