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

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13 Nov 2007, 09:20

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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?

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.

Last edited by nevergiveup on 13 Nov 2007, 10:29, edited 1 time in total.

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.

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).

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.

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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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

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?

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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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.

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

Let me know. Cheers!
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