Probability

[eskander]

Could you please help me out. I must be missing some step in calculations...

There are 10 bottles wine in a carton. Three bottles will be selected and examined at random. If any of the three bottles is defective, the carton will be sent back. If a carton contains exactly three defective bottles, what is the probability that the carton will not be sent back?
(A) 1/9
(B) 3/10
(C) 7/10
(D) 7/24
(E) 17/24

[coolhabhi]

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

[ihatemaths]

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

[Yermek1984]

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

[eskander]

btw OA is 7/24