Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 5

shashank_prabhu last edited by
CAT 100%iler, 5 times AIR 1, Director  Learningroots, Ex ITC, Pagalguy, TAS
AE:FE=2:3
AB:DF=2:3
A(ABE)/A(DEF)=(AB/DF)^2=4/9
A(DEF)=4.5
A(BDF)=7.5
A(FBCD)=15
A(AECD)=5.5

@shashank_prabhu @zabeer mechanic has to check the first two machines only if he is able to identify 2 good or 2 defective machines, he can catch his bus; otherwise he cant. So we have to consider the cases with only first 2 machines; that gives me gg, dd, gd, dg ie 4 total cases and not 6. My answer is 2/4. Can you please explain why we have to consider the arrangements of ddgg ?

@VikrantGarg
Probability of catching the bus = Probability of getting faulty machines in the first 2 tests (this part is clear right?)
This is possible in 2 ways  if we get DD or if we get GG as the result of first 2 tests.
In your doubt, if we get DG or GD we will require to do one more test, which would make the mechanic miss the bus. So DG and GD are not favourable cases.
So total number of ways = 4!/(2! x 2!) = 6 ways.
Favourable cases = 2
Probability = 2/6 = 1/3
Is it clear ?