Q#1: The seven dwarves (of Snow White fame) have seven beds in a dormitory room. One night, Sleepy is so tired that he falls asleep on the first bed that he sees. This has happened before, so the other dwarves follow a standard procedure as they enter the dormitory room later that night: if a dwarf can sleep in his own bed, he does so, but if not, he selects a random bed and uses that one. On the night in question, Grumpy is the fourth dwarf to enter the room to go to bed. What’s the probability that he sleeps in his own bed?
By the end D2 has occupied the bed, whether it is his or somebody else's, the bed of D2 has been occupied because if his bed were available when he entered, he would have occupied it himself.
Same logic holds good for D3 as well, that is, by the end D3 has occupied the bed, the bed of D3 has been occupied either by D3 himself or by somebody else.
This means that when D4 aka Grumpy enters the room, the beds of D2 and D3 have already been occupied out of the 3 occupied beds. If 1 occupied bed is of D4, he won't be able to occupy his bed.
Probability that the bed of D4 was occupied before he entered the room = Probability that D4 is unable to occupy his own bed = \(\overline {P(D4)} = \frac{1}{{7 - 2}} = \frac{1}{5}\)
Probability that D4 is able to occupy his own bed = \(P(D4) = 1 - \frac{1}{5} = \frac{4}{5}\)
Thankyou Sir....thankyou very much for the different approach