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:

Question

Infinity Learn · Accepted Answer

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)

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:

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