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(A) The total number of cards is $20$. Number of cards with $I = 10$. Number of cards with $T = 10$. Since the cards are drawn with replacement,the pr

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$1/8$

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(A) The total number of cards is $20$. Number of cards with $I = 10$. Number of cards with $T = 10$. Since the cards are drawn with replacement,the probability of drawing $I$ in any single draw is $P(I) = \frac{10}{20} = \frac{1}{2}$. The probability of drawing $T$ in any single draw is $P(T) = \frac{10}{20} = \frac{1}{2}$. To form the word $IIT$,we need to draw $I$ first,$I$ second,and $T$ third. Since the events are independent due to replacement,the probability is $P(I) 	imes P(I) 	imes P(T) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$.

You are given a box containing $20$ cards. Out of these,$10$ cards have the letter $I$ printed on them,and the other $10$ cards have the letter $T$ printed on them. If you draw three cards one after another with replacement,what is the probability of forming the word $IIT$?

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