In a lot of 12 washing machines , there are 3 defective pieces . A person has ordered 4 washing machines . Find the probability that all the four are good .

My working :

X-B(4 , 0.75)
P(X=4) = 4C4(0.75)^4 (0.25)^0 = 81/256

But my answer is wrong . Where is my mistake again ? Thanks

X is not binomial because you're choosing without replacement (in fact, X is hypergeometric).

A combinatorial approach (which is what the hrpergemetric distribution is actually based on):

.

Do you see where each bit comes from?

__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  

As Mr. Fantastic (and he really is!) said, this is sampling without replacement and the formula you used applies only to sampling with replacement.

It's better to think like this: there are, to start with, 12 machines, 9 of which are good. The probability that the first machine selected is good is 9/12= 3/4. Assuming that happens, there are now 11 machines, 8 of which are good. The probability that the second machine selected is good is 8/11. Assuming that happens, there are now 10 machines, 7 of which are good. The probability the third machine is selected is 7/10. Finally, we have 9 machines left, 6 of which are good and the probability that the fourth machine selected is good is 6/9= 2/3. We can assume each selected was good because we are only calculating the probability that all four selected are good: it is (9/12)(8/11)(7/10)(6/9)= (3/4)(8/11)(7/10)(2/3)

Oh oh , i know my mistake . Thanks a lot , Mr F and Hallsofivy for that effort .