i think a,b,d, are true
if first and third are mobile while second is not : – – –
5/15 10/14 4/13
in third position 4/13 is because we have used one mobile in first position,and rest 10 which are not mobile in second out of total 14 because we used 1 mobile previous and so on for 3rd position
option B…
if first four are wired let suppose position are – – – –
4/15 3/14 2/13 1/12
total wired are 4 and first position fill out of 4 and total are 15 now when we used 1st wired rest total remaining 13, and rest wired remaining 3 so 2nd position filled 3 out of 14 and so on
D
the probability of selecting two phone it is simple just follow option b