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

Question

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

Bunuel · Accepted Answer

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.

aliensoybean · Answer

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.

nevergiveup · Answer

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

bmwhype2 · Answer

why isnt it 0.3*0.3*0.3*0.3*0.7*0.7  x 6C4

nevergiveup · Answer

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.

pmenon · Answer

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

GMAT TIGER · Answer

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.

walker · Answer

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

gmatraider · Answer

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

???

GMATBLACKBELT · Answer

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.

incognito1 · Answer

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

bhatia_ash2002 · Answer

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?

incognito1 · 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)

bhushan252 · Answer

out of 4 visitors => 4 buys & 2 dont buy => 0.3^4 x 0.7^2 => 0.3969%

sunny4frenz · Answer

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 :-)

srivas · Answer

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

jeeteshsingh · Answer

6c4 x 0.3^4 x 0.7^2 = 15 x 0.0081 x 0.49 = 0.059

vonshuriz · Answer

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.

Capricorn369 · Answer

P(at least one will not buy) = 1 - P(Everybody Buys)
 = 1-0.7^6

Let me know.
Cheers!

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

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