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A store buys 10 bottles of alcohol, including 7 bottles of [#permalink]
04 Nov 2007, 13:41

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45% (medium)

Question Stats:

63% (03:06) correct
36% (02:23) wrong based on 268 sessions

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.

Re: combination PS [#permalink]
04 Nov 2007, 13:55

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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

Last edited by beckee529 on 04 Nov 2007, 14:13, edited 1 time in total.

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

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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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

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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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).

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!}.

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.

Re: combination PS [#permalink]
25 Jun 2012, 21:25

1

This post received KUDOS

Expert's post

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]
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]
10 Oct 2013, 19:31

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