Probability

Let the 1st digit be prime. Then, the total number of selections = 4*4*3*2
Let the 2nd digit be prime. Then, the total number of selections = 4*4*3*2
Let the 3rd digit be prime. Then, the total number of selections = 4*4*3*2
Let the 4th digit be prime. Then, the total number of selections = 4*4*3*2
Total number of such cases = 4*4*4*3*2

Total number of combinations to form a 4 digit code(If 0 can be placed first and if repetition is not allowed) = 10*9*8*7

Probability = (4*4*4*3*2)/(10*9*8*7) = (4*2)/(5*3*7)= 8/105

p.s: One of the few Probability problems I solved in my life

Hi Ganesh

Can u please post the OA.

This is how I tried it:

ways of selecting prime = 4C1
ways of selecting composite = 6C3
total ways of selecting any = 10C4

P(Final) = (4C1 * 6C3)/ 10C4 = 8/21

can anyone explain the question first rather than the answer

though i am not bad with probability
this question didn make sense