A store buys 10 bottles of alcohol, including 7 bottles of

Question

A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought.

A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

Show Answer

Official Answer

D

A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

Bunuel · Accepted Answer

We can solve the way you propose as well, with a little correction. A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought. A. 2/5 B. 3/5 C. 2/3 D. 1/2 E. 4/7 We have 7 whiskey (W) and 3 non-whiskey bottles and want to find the probability of selling WWWWNN (4 whiskey bottles and 2 non-whiskey bottles). P=\frac{6!}{4!2!}*(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}*\frac{4}{7})*(\frac{3}{6}*\frac{2}{5})=\frac{1}{2} . We are multiplying by \frac{6!}{4!2!} as scenario WWWWNN can occur in # of ways: NNWWWW (first bottle sold was non-whiskey, second bottle sold was non-whiskey, third bottle sold was whiskey, ...); NWNWWW; NWWNWW; ... Some # of combinations, which basically equals to # of permuations of 6 leeters WWWWNN out of which 4 W's and 2 N's are identical --> \frac{6!}{4!2!} . Answer: D. Similar problems: viewtopic.php?f=140&t=96468&p=742767&hilit=+google#p742767 viewtopic.php?f=140&t=56812&hilit=+probability+occurring+times viewtopic.php?f=140&t=88069&hilit=+probability+occurring+times viewtopic.php?f=140&t=87673&hilit=+probability+occurring+times ps-probability-teams-68186.html?hilit=occur%20permutations%20letters#p728696 probability-question-100222.html?hilit=occur%20permutations%20letters#p772756 Hope it's clear.

beckee529 · Answer

please confirm if i am correct

10 bottles - 7 whiskeys / 3 other

6 sold, disregard 4

probability: F/T

T:
total possible: 10C6 = 10!/6!4! = 30*7 = 210

F:
selling 4 whiskeys: 7C4 = 7!/4!3! = 35
the rest will be  2  so: 3C2 = 3!/2!1! = 3
multiplying the two together : 35*3 = 105

the probability: F/T = 105/210 = 1/2 D

edit sorry to confuse i just updated

shubhampandey · Answer

I understand the answer. 

However something is confusing me when I look at it second time.

The only way 6 bottles are sold is if, We sell 

1. 6 whisky
2. 5 whisky 1 Non whisky
3. 4 whisky and 2 non whisky
4. 3 whisky and 3 non whisky.


so total possible way of selling 6 bottles = 4
so total possible number when exactly 4 whiskys are sold = 1

so probabilty = 1/4

what is wrong with this equation??

MBAwannabe10 · Answer

Can someone explain to me why the following won't work? 
i did binomial approach 
P (of getting whiskey among the 10 bottles) = 7/10
P(of getting something else) = 3/10

so getting 4 whiskey out of 6: 
C (6,4)*(7/10)^4*(3/10)^2

but this doesn't give me 1/2 for an answer? 
help?

mainhoon · Answer

I guess this is simply 7C4 3C2 / 10C6 = 1/2?

KarishmaB · Answer

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.

luckyme17187 · Answer

Hi All, Please let me know whether my approach is correct.. 7C4*3C2/ 10C6 = 1/2 .

7C4- Out of 7 Whiskey bottles 4 sold 3C2 -Out of remaining 3 bottles 2 sold 10C6-- Out of 10 bottles - 6 sold.

KarishmaB · Answer

You have used combination approach which is fine too.

KarishmaB · Answer

Responding to a pm:

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)

mmdfl · Answer

If 5 or 6 bottles of whisky were sold, then it is a true statement that 4 bottles was sold (and 3, and 2, and 1). I was holping for and "selling exactly 4 whiskeys"... I always get confused on this...

Bunuel · Answer

No, selling 4 bottles means specifically selling exactly 4 bottles. It does not include scenarios where 5 or 6 bottles are sold. In this context, "selling 4" is understood as "selling exactly 4," not at least 4. If the scenario were to include selling 4 or more bottles, the question would typically specify "selling at least 4 whiskeys."

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