There are 2 possibilities :
a) We have 2 chains ; one of length 3 and another of length 1 ..
For this we select 1 slot out of the 8 slots for the 3 keys and the one key remaining can be placed in one of remaining 7 blocks..But the keys which are being chosen to form a chain of 3 also matters..
So probability of doing so : 8C1 * 4C1 * (1/8)3 * (7/8)
= 8 * 4 * 7 / (84)
= 28 / 83
b) We have only one chain of length 4 :
So probability of doing so : 8 * (1/8)4 [As we can have 4 keys into any of the 8 slots , so we are multiplying by 8]
= 1 / 83
Therefore P(at least a chain of size 3) = 28 / 83 + 1 / 83
= 29 / 83
= 29 / 512
= 0.057 (correct to 3 decimal places)