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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 

 

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