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

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A store buys 10 bottles of alcohol, including 7 bottles of  [#permalink]

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New post 04 Nov 2007, 13:41
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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
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Re: combination PS  [#permalink]

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New post 15 Sep 2010, 10:26
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MBAwannabe10 wrote:
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?

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.

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Hope it's clear.
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Re: combination PS  [#permalink]

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New post Updated on: 04 Nov 2007, 14:13
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gowani wrote:
please explain...I'm new to combinations and need help setting it up. Thanks in advance.

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


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

Originally posted by beckee529 on 04 Nov 2007, 13:55.
Last edited by beckee529 on 04 Nov 2007, 14:13, edited 1 time in total.
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  [#permalink]

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New post 04 Nov 2007, 14:06
yes, you are correct.

what do you mean by:

the rest will be 4 so: 3C2 = 3!/2!1! = 3


also, what does F and T mean?

thanks for your explanation
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  [#permalink]

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New post 04 Nov 2007, 14:11
gowani wrote:
yes, you are correct.

what do you mean by:

the rest will be 4 so: 3C2 = 3!/2!1! = 3


also, what does F and T mean?

thanks for your explanation


the stem tells us that there were 6 sold and the rest are non whiskey alcohol. the "rest" is 4 so we do not consider those. Since they ask about 4 bottles of whiskey sold from the 6, 6-4 = 2 is the number of combinations for the 3 non whiskey bottles

F = favorable
T = total possibilities

those are the heart and core of probability that you have to fundamentally know to understand probability

for instance, 1/2 probability for head/tail coin flip

i hope that makes sense, good luck!
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  [#permalink]

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New post 26 Nov 2007, 19:59
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how did you quickly derive that 10!/6!4! = 30*7 ?
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confusion  [#permalink]

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New post 01 Dec 2007, 13:04
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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??
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Re: combination PS  [#permalink]

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New post 21 Sep 2008, 10:01
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This is a classic hypergeometric distribution problem.
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Re: combination PS  [#permalink]

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New post 21 Sep 2008, 22:41
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oziozzie wrote:
This is a classic hypergeometric distribution problem.

Whats that ??

gowani wrote:
please explain...I'm new to combinations and need help setting it up. Thanks in advance.
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


I agree with above explanations.

Total outcomes : 10C6
Favourable Outcomes : 7C4*3C2

Probability : 7C4*3C2/10C6 = 105/210 =1/2
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Re: combination PS  [#permalink]

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New post 28 Sep 2009, 10:10
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

Soln: Probability that 4 bottles of whisky will be sold is
= 7C4 * 3C2/10C6
= 1/2

Ans is D
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Re: combination PS  [#permalink]

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New post 01 Oct 2009, 21:50
Solution:

C(6,4) / C(10,6)
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Re: combination PS  [#permalink]

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New post 15 Sep 2010, 09:55
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?
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Re: combination PS  [#permalink]

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New post 15 Sep 2010, 10:06
The binomial approach only works if the probability of outcomes of each successive event remains constant

This is not true here

After the first bottle is sold, the probability of whisky and non-whisky changes.

Posted from my mobile device
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Re: combination PS  [#permalink]

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New post 15 Sep 2010, 16:21
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I guess this is simply 7C4 3C2 / 10C6 = 1/2?
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Re: combination PS  [#permalink]

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New post 21 Sep 2010, 17:18
combinations are killing me! Great explanation Beckee
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Re: combination PS  [#permalink]

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New post 11 Oct 2010, 04:21
gowani wrote:
A store buys 10 bottles of alcohol,

Bag of 10 choices.

gowani wrote:
including 7 bottles of whiskeys.

Bag is made of 7 whiskey and 3 nonwhiskey.

gowani wrote:
In the evening, 6 bottles of alcohol are sold one by one

Combo box arrangement
(_)(_)(_)(_)(_)(_)/6!

gowani wrote:
and the rest is consumed by the personnel.

um...ok. drinkdrinkdrinkdrinkdrink

gowani wrote:
What is the probability of selling 4 whiskeys among the 6 bottles?

Probability Table: Create and work backwards.

# of Whiskey: Events
0:
1:
2:
3:
4:
5:
6:
-------------------------------------
Total =

*******************************************************

Total = 10 bottles pick 6 = 10C6 = 10*9*8*7*6*5/6*5*4*3*2*1 = 210

6 Whiskey: 7 whiskey pick 6 = 7C6 = 7 <---------don't need this, but good for practice.

5 Whiskey: 7 whiskey pick 5 AND 3 non whiskey pick 1 = 7C5 * 3C1 = 63 <----------don't need this either.

4 Whiskey: 7 whiskey pick 4 AND 3 non pick 2 = 7C3 * 3C2 = 105 <-----This is what we need. I'll finish the table for fun.

3 Whiskey: 7 whiskey pick 3 AND 3 non pick 3 = 7C3 * 3C3 = 35
OR sum the table total: 210 - (7+63+105) = 35

2 Whiskey: 7 whiskey pick 2 AND 3 non pick 4 = impossible = 0

1 Whiskey: also impossible = 0

0 Whiskey: also impossible = 0

*******************************************************

# of Whiskey: Events
0: 0
1: 0
2: 0
3: 35
4: 105
5: 63
6: 7
-------------------------------------
Total = 210

*******************************************************

Use the info in the table to answer any probability question.
P(Whiskey = 4) = 105/210 = 1/2
P(Whiskey = prime number) = (0+35+63)/210
P(Whiskey > 3) = (105+63+7)/210

gowani wrote:
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



ANS: D
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Re: combination PS  [#permalink]

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New post 25 Jun 2012, 21:25
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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.
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Re: A store buys 10 bottles of alcohol, including 7 bottles of  [#permalink]

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New post 24 Sep 2012, 22:55
Total Number of Alcohol Bottles = 10
Whiskey Bottles = 7

No of bottles sold by the Store = 6

To calculate the probability that 4 out of these 6 is whiskey we will use the following formula : -


P (A) = Total No. of Cases favorable to the happening of A
____________________________________________

Total No of exhaustive equally likely cases

So the denominator becomes , C (10,6) which = 210

To find the number of combinations we use :

C (7,4) x C (3,2) ie Four out of the 7 whiskey bottles makes up the four bottles (of the 6 sold) and 02 of the remaining non whiskey bottles make up the remaining two ...

This is equal to 105 ....

Dividing to get the probability we get 105/210 = 1/2 (D)
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Re: A store buys 10 bottles of alcohol, including 7 bottles of  [#permalink]

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New post 12 Nov 2013, 10:28
One way I thought you can look at the question is...

what is the probability of 4 whiskeys and 2 bottles of alcohol being sold from a collection of 10 bottles if one bottle is being sold at a time.

Thus the probability would be as follows:

(7/10)*(6/9)*(5/8)*(4/7)*(3/6)*(2/5) = 1/30 => this is the probability that 4 whiskeys and 2 alcohol bottles are sold out of 10 bottles.

but bottles can be sold in any order, so considering all orders in which the bottle may be sold that => 6!/(4!*2!) = 15

multiplying both, we get

15/30 => 1/2

I do hope that this method is right and hope that somebody benefits from this solution.
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Re: A store buys 10 bottles of alcohol, including 7 bottles of  [#permalink]

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New post 15 Aug 2014, 10:35
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.
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Re: A store buys 10 bottles of alcohol, including 7 bottles of   [#permalink] 15 Aug 2014, 10:35

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