One of the ten available keys opens the door. If we try the keys one after another, then find the following
(i) the probability that the door is opened in the first attemt.
(ii) the probability tht the door is opened in the second attempt.
(iii) the probability that the door is opened in the third attempt.
(iv) the probability that the door is opened in the tenth attempt.

Explanation

Let `A_(i)` be the event that door is opened in ith attempt.
(i) Since our of 10 keys only one key is correct, the probability that the door is opened in the first attempt is same as probability that the door is opened with randomly chosen key.
`So, P(A_(1))=1/10.`
(ii) If door is opened in the second attempt, then the fiest attempt must have failed.
So, P (door is opened in second attempt)
`=P(A_(1)'nnA_(2))=P(A_(1)')P(A_(2)//A_(1)')`
`P(A_(1)')=P("first attempt is fail")=9/10`
`P(A_(2)//A_(1))=P` (second attempt is success when it is known that first attempt is fail)= P(door is opened with randomly chosen key `=1/9` from 9 keys)
So, `P(A_(1)'nnA_(2))=P(A_(1)')P(A_(2)//A_(1))=9/10xx1/9=1/10`
(iii) P (door is opened in third attemt)
`=P(A_(1)'nnA_(2)'nnA_(3))`
`=P(A_(1)')P(A_(2)'//A_(1))P(A_(3)//(A_(2)'nnA_(1)))`
`=9/10xx8/9xx1/8=1/10`
(iv) P (door is opened in tenth attempt)
`=9/10xx8/9xx7/8xx6/7xx5/6xx4/5xx3/4xx2/3xx1/2xx1= 1/10`