In a box, there are 20 cards, out of which 10 are labelled as A and the remaining 10 are labelled as B. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third B-card is:

The possibilities are (AA),(A,B,A),(A,B,B,A) The probability of getting AA is 12·12=14  The possibilities to the second choice (A,B,B,A) are (A,B,B,A) or (B,A,B,A),(B,B,A,A),(A,B,B,A) The probability of getting (A,B,B,A) is 3·12·12·12·12=316 Therefore, the required probability will be 14+14+316=1116 Hint: The possibilities that the second A-card appears before the third B-card are (AA),(A,B,A),(A,B,B,A)