IMO : D

there are exactly three defective bottles and 7 non defective bottles.


So if the three bottles are picked up from the 7 non defective bottles then the case will not be sent back.

=>probability of choosing 1st non defective bottle = 7/10
=>probability of choosing 2nd non defective bottle = 6/9
=>probability of choosing 3rd non defective bottle = 5/8

=> (7*6*5)/(10*9*8)
=> 7/24

defective->3
non-defe->7
total->10

condition for returning back is atleast one i.e if even 3c1 package is sent back

condition ensuring non-return is : (7c3 * 3c0)/10c3 = 7/24

coolhabhi, I must disagree with u. May be I was wrong on this, but here is my option:
1) the probability of choosing at least 1 defective bottle = (3*7*6)/(10*9*8)= 7/40;
2) the probability of choosing at least 2 defective bottles = (3*2*7)/(10*9*8) = 7/120;
3) the probability of choosing at least 3 defective bottles = (3*2*1)/(10*9*8) = 1/120;

Since, we have to figure out the probability when a cartoon won't sent back, then the P. = (1-(7/40+7/120+1/120)) = 91/120;

I'm not sure, but there are no correct answers. Correct me plz if I'm not right.

Yermek

btw OA is 7/24