The Student Room Group

S1 MEI Q10 difficulty

I cannot find the solution to Q10 part (iv) from the EMI S1 text book p136.

A hotel caters for business clients who make short stays. Past records suggest that the probability of a randomly chosen client staying X nights in succession is as follows:
r 1 2 3 4 5 6+
P(X=r) 0.42 0.33 0.18 0.05 0.02 0

(i) Draw Distribution. [No problem]
(ii) Find Mean & S.D. [No problem] ans. 1.92 & 0.987
(iii) Find the probability that a randomly chosen client who arrives on Monday will still be in the hotel on Wednesday night. [No problem] ans. 0.25
(iv) Find the probability that a client who has already stayed two nights will stay for at least one more night. [Big problem!!] Book answer gives 0.431
Reply 1
P(X >= 3 | X >= 2)
= P(X >= 3 and X >= 2) / P(X >= 2) . . . . . definition of conditional probability
= P(X >= 3) / P(X >= 2) . . . . . since X >= 3 implies X >= 2
= (0.18 + 0.05 + 0.02) / (1 - 0.42)
= 0.431