VeritasPrepKarishma wrote:
Responding to a pm:
We cannot use Binomial here. Think 'flipping a coin' or 'with replacement' when you want to use binomial.
You are not replacing the bottles with identical bottles every time one of them is sold/consumed. Hence, replacement is not taking place here. The probability of selling a whiskey bottle changes after you sell one. Therefore, the probability does not stay at 7/10 so you cannot use binomial.
You have 7 W and 3 N.
So probability of selling 4 W and 2 N = (7/10)*(6/9)*(5/8)*(4/7)*(3/6)*(2/5)* 6!/4!*2!
Now, as for your question: "A box contains 100 bulbs out of which 10 are defective. A sample of 5 bulbs is drawn.The prob. that none is defective is ?"
You need to give the exact question. Did they mention whether they are replacing the bulbs after each draw? If the solution uses binomial, when you draw a bulb, you need to replace it and then draw another one. Otherwise, you cannot use binomial here.
Responding to a pm:
Quote:
I see that a lot of people used the combination formula for this problem: (7C4 * 3C2) / 10C6. However, even though you get the "right" answer using this approach, this formula is not accounting for the number of combinations that you get out of the WWWWNN sample (i.e. WNWNWW, WWNNWW, WWNWNW, etc.)
Having said this, i dont think it is appropiate to use this formula for this problem. DO you have any thoughts about this? Thanks.
Yes, it is using the combination approach for probability. It is alright to use it because you are not arranging in both numerator and denominator. Even if you do, note that you have 10 distinct bottles. Once you select 6 bottles, the number of ways of arranging them is 6!.
So you get (7C4 * 3C2 * 6!) / (10C6 * 6!)
This is the same as 7C4*3C2 / 10C6 (the 6! anyway gets cancelled out)