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The probability that a visitor at the mall buys a pack of

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Re: the prob that a visitor at the mall buys a pack of candy is [#permalink]

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New post 24 Feb 2012, 13:20
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fortsill wrote:
where's the right answer posted? how do i know if i'm right?


I added the correct answer under the spoiler in the initial post. Below is a solution.

The probability that a visitor at the mall buys a pack of candy is 0.3. If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

Since the probability that a visitor buys a candy is 0.3, then the probability that a visitor DOES NOT buy a candy is 1-0.3=0.7.

We want the probability that exactly 4 visitors out of 6 will buy a candy, so the probability of BBBBNN (where B denotes a visitor who buys a candy and N denotes a visitor who does not buy a candy). Each B has the probability of 0.3 and each N has the probability of 0.7, so we have \(0.3^4*0.7^2\).

Next, BBBBNN case can occur in # of different ways: NNBBBB, NBNBBB, NBBNBB, ... (first two visitors doesn't buy and next four does; first doesn't buy, second does, third doesn't and next three does; ...) Basically it's # of permutations of 6 letters BBBBNN out of which 4 B's and 2 N's are identical, so \(\frac{6!}{4!*2!}\).

Finally \(P(B=4)=\frac{6!}{4!*2!}*0.3^4*0.7^2\).

Check the following links for different scenarios similar problems:
the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-55689.html
probability-85523.html

Hope it helps.
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Re: The probability that a visitor at the mall buys a pack of [#permalink]

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New post 12 Mar 2017, 05:06
bmwhype2 wrote:
The probability that a visitor at the mall buys a pack of candy is 0.3. If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?


CORRECT ANSWER:
[Reveal] Spoiler:
\(\frac{6!}{4!*2!}*0.3^4*0.7^2\)



Why do we have to multiply \(\frac{6!}{4!*2!}\) ?
I don't think the sequence in which people buy matters here. Does it ?

Bunuel, kindly help.
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Re: The probability that a visitor at the mall buys a pack of [#permalink]

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New post 12 Mar 2017, 06:22
Shruti0805 wrote:
bmwhype2 wrote:
The probability that a visitor at the mall buys a pack of candy is 0.3. If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?


CORRECT ANSWER:
[Reveal] Spoiler:
\(\frac{6!}{4!*2!}*0.3^4*0.7^2\)



Why do we have to multiply \(\frac{6!}{4!*2!}\) ?
I don't think the sequence in which people buy matters here. Does it ?

Bunuel, kindly help.


Please read the whole thread before posting a question: https://gmatclub.com/forum/the-probabil ... l#p1049555
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The probability that a visitor at the mall buys a pack of [#permalink]

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New post 29 Mar 2017, 19:40
Can someone help me understand why we don't consider the combinations
So the probability is BBBBNN(B-buy N-not buying)
Therefore p(B)=3/10 p(N)=7/10
3/10^4 * 7/10 ^2 * 6!/4!*2!

Is this also the correct answer??
Re: The probability that a visitor at the mall buys a pack of   [#permalink] 29 Mar 2017, 19:40

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