Bag contains 5 red balls and 7 white balls. So the total number of all favorable cases is n(S) = 5 + 7=12
Let A represents first ball as white ball, and B be second ball as white ball.
Then the probability of drawing two white balls without replacement is
P(2 white balls without replacement)
= P(A)P(B|A)
= \(\cfrac7{12}\times\cfrac{6}{11}\) (as there are 7 white balls in first draw out of 12 balls, and 6 white balls in second draw out of 11 balls as the balls are not replaced)
Hence the required probability is \(\cfrac7{22}\)