Probability that both mango is rotton = 4c2/10c2 [ i.e. 4 are rotton u choose any two of them in 4c2 ways]
Probability that atleast of of them is good = Total [since total probability is 1] - Probability that both mango is rotton
= 1 - 4c2/10c2
= 13/15
Probability that both mango is good = 10 - 4 = 6 choose any 2 mango out of these 6 = 6c2/ 10c2 =1/3
So required probability [ Both are good ] = 15/13 * 1/3
= 5/13 [BAYES THEOREM p(B/A) = p(A intersection B) /P(A) ]