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 3^{rd} 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 1^{st} wired rest total remaining 13, and rest wired remaining 3 so 2^{nd} position filled 3 out of 14 and so on

D

the probability of selecting two phone it is simple just follow option b