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

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SVP
Joined: 21 Jan 2007
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The probability that a visitor at the mall buys a pack of  [#permalink]

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Updated on: 24 Feb 2012, 12:21
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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?

$$\frac{6!}{4!*2!}*0.3^4*0.7^2$$

Originally posted by bmwhype2 on 13 Nov 2007, 08:20.
Last edited by Bunuel on 24 Feb 2012, 12:21, edited 4 times in total.
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Re: the prob that a visitor at the mall buys a pack of candy is  [#permalink]

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24 Feb 2012, 12:20
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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:
probability-85523.html

Hope it helps.
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13 Nov 2007, 08:39
6 people and probability of buy candy 0.3

probability of exact 4 buys candy will be 0.3*0.3*0.3*0.3*0.7*0.7=0.003969
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13 Nov 2007, 08:42
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eileen1017 wrote:
6 people and probability of buy candy 0.3

probability of exact 4 buys candy will be 0.3*0.3*0.3*0.3*0.7*0.7=0.003969

why isnt it 0.3*0.3*0.3*0.3*0.7*0.7 x 6C4
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Updated on: 13 Nov 2007, 09:29
Ok, this is just like toss the coin three times what is the probability that exact two will land on tail. Since each coin has 0.5 to be head and 0.5 to be tail. The probability will be 0.5(for land on tail)*0.5(for land on tail)*0.5( for land on head). This question has the same logic.

Never mind. My example is wrong. The probability of the coin will be 3/8.

Originally posted by nevergiveup on 13 Nov 2007, 09:06.
Last edited by nevergiveup on 13 Nov 2007, 09:29, edited 1 time in total.
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13 Nov 2007, 09:09
i would have thought that it will be 6C4 , which gives the number of ways 4 people out of 6 can buy candy.

Then, take that number and multiply it by 0.3
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13 Nov 2007, 22:31
bmwhype2 wrote:
the prob that a visitor at the mall buys a pack of candy is .3.
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

p = 0.3^4 0.7^2

it cannot be multiples by 6c4 because the visitors are not going to have 4 buy andf 2 not-buy 6c4 times. we know for sure that it is going to happen only once.

therefore, p = 0.3^4 0.7^2 is correct.
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14 Nov 2007, 01:36
4
Here is how i approached it:

1. What is a prob of a simple sequence where the first 4 people were to buy so (0.3)^4 * (.7)^2

2. But this is also a combo problem because out of 6 people, any groups of 4 can be chosen to buy--not just the first 4. In another words, there are 6C4 ways to rearrange (0.3)(0.3)(0.3)(0.3)(0.7)(0.7).

So, you have to multiply by 6C4.
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23 Nov 2007, 23:09
a few theory statements....

p,q -probability, where p+q=1.
n - number of events.

Thus,
1^n=(p+q)^n=C(0,n)*p^n+C(1,n)*p^n*q+...+C(m,n)*p^(n-m)*q^m+...+C(n,n)*q^n

in our case:

P=C(m,n)*p^(n-m)*q^m, where m=4, n=6, p=0.7, q=0.3

Therefore,
C(4,6)*0.3^4*0.7^2
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27 Jan 2008, 11:24
CaspAreaGuy wrote:
could it be (1-0.7*.03^5)*6?

Why have you subtracted the .07 from 1? That would simply make your equation (.3^6)*6. I was thinking (.3^5*.7)*6

???
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27 Jan 2008, 22:21
bmwhype2 wrote:
the prob that a visitor at the mall buys a pack of candy is .3
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

I also came across the same dilema here to multiply by 6C4 or not.

3/10*3/10*3/10*3/10*7/10*7/10

Now obvs we could have 6!/4!2!

Is this a problem you made up or is there an OA? Id like to see whether this is the correct approach. I went for multiplying by 6C4.

B/c there is no restriction on the order here. So we can have several different orders.
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29 Jan 2008, 13:53
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bmwhype2 wrote:
What is the P(at least one will not buy)?

Why is this not 1 - (0.3^6)?

Reason:
The probability that atleast one will not buy = 1 - (probability that everyone buys)
where (probability that everyone buys) = 0.3^6

(0.3^5) * (0.7) * 6 represents the probability that EXACTLY one person does not buy
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29 Jan 2008, 16:39
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To the question:

What is the P(at least one will not buy)?

Cant we just rephrase it as:

What is the P(no one will buy)?

Which can be solved as

1-0.7^6

Why is this not the correct approach and answer?
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29 Jan 2008, 16:43
1
bhatia_ash2002 wrote:
To the question:

What is the P(at least one will not buy)?

Cant we just rephrase it as:

What is the P(no one will buy)?

Which can be solved as

1-0.7^6

Why is this not the correct approach and answer?

0.7^6 is the probability that no one will buy
1 - (0.7^6) is hence the probability that atleast one person will buy, as opposed to the probability that atleast one person will not buy (as required by the question)
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25 Jul 2009, 10:06
out of 4 visitors => 4 buys & 2 dont buy => 0.3^4 x 0.7^2 => 0.3969%
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21 Sep 2009, 00:55
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Here is the brutal way of verifying the result ....

I think answer should be 6C4*(0.3)^4*(o.7)^2

EXAMPLE : Suppose we have 3 visitors ....
then complete tree would be like :

0.3*0.3*0.3 = 0.027 [ when all r gng for shopping]
0.3*0.3*0.7 = 0.063 [ 1&2 goes for shopping]
0.3*0.7*0.3 = 0.063 [ 1&3 goes for shopping]
0.7*0.3*0.3 = 0.063 [ 2&3 goes for shopping]
0.3*0.7*0.7 = 0.147 [ only 1 goes for shopping]
0.7*0.3*0.7 = 0.147 [ only 2 goes for shopping]
0.7*0.7*0.3 = 0.147 [ only 3 goes for shopping]
0.7*0.7*0.7 = 0.343 [ no one goes]

Total = 1

clearly If multiply by nCr ... only then we can get the correct answer

Consider for Kudos ....if u like it
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27 Sep 2009, 01:35
the prob that a visitor at the mall buys a pack of candy is .3
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

Soln:
6C4 * (.3)^4 * (.7)^2
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14 Feb 2010, 09:03
bmwhype2 wrote:
the prob that a visitor at the mall buys a pack of candy is .3
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

6c4 x 0.3^4 x 0.7^2 = 15 x 0.0081 x 0.49 = 0.059
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04 Mar 2010, 04:07
incognito1 wrote:
bhatia_ash2002 wrote:
To the question:

What is the P(at least one will not buy)?

Cant we just rephrase it as:

What is the P(no one will buy)?

Which can be solved as

1-0.7^6

Why is this not the correct approach and answer?

0.7^6 is the probability that no one will buy
1 - (0.7^6) is hence the probability that atleast one person will buy, as opposed to the probability that atleast one person will not buy (as required by the question)

You are right!! The question asks for the probability that at least 1 does not buy. That means that everything has to be considered from 1 buying nothing to all 6 people buying nothing. Therefore the right solution should be 1 - p(everybody buys) = 1 - 0.3^6

And if everybody is buying the order does not matter in this respect.
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12 Sep 2011, 15:25
bmwhype2 wrote:
What is the P(at least one will not buy)?

= 1-0.7^6

Let me know.
Cheers!
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Re:   [#permalink] 12 Sep 2011, 15:25

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