A woman has 10 keys,out of which only one ope | vedclass

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(D) The total number of keys is $10$,and only $1$ key opens the lock.Since the keys are tried one after another without replacement,the probability th

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$\frac{1}{10}$

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(D) The total number of keys is $10$,and only $1$ key opens the lock.Since the keys are tried one after another without replacement,the probability that the $k$-th key works is the probability that the first $k-1$ keys fail and the $k$-th key succeeds.Let $F_i$ be the event that the $i$-th key fails and $S_i$ be the event that the $i$-th key succeeds.We want to find $P(F_1 \cap F_2 \cap F_3 \cap F_4 \cap F_5 \cap F_6 \cap S_7)$.Using the multiplication rule of probability:$P(F_1) = \frac{9}{10}$$P(F_2 | F_1) = \frac{8}{9}$$P(F_3 | F_1 \cap F_2) = \frac{7}{8}$$P(F_4 | F_1 \cap F_2 \cap F_3) = \frac{6}{7}$$P(F_5 | F_1 \cap F_2 \cap F_3 \cap F_4) = \frac{5}{6}$$P(F_6 | F_1 \cap F_2 \cap F_3 \cap F_4 \cap F_5) = \frac{4}{5}$$P(S_7 | F_1 \cap F_2 \cap F_3 \cap F_4 \cap F_5 \cap F_6) = \frac{1}{4}$Multiplying these probabilities:$P = \frac{9}{10} 	imes \frac{8}{9} 	imes \frac{7}{8} 	imes \frac{6}{7} 	imes \frac{5}{6} 	imes \frac{4}{5} 	imes \frac{1}{4} = \frac{1}{10}$.

$A$ woman has $10$ keys,out of which only one opens a lock. She tries the keys one after another (keeping aside the failed ones) until she succeeds in opening the lock. What is the probability that the seventh key works?

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