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# In a department store prize box, 40% of the notes give...

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VP
Joined: 06 Sep 2013
Posts: 1491
Concentration: Finance
In a department store prize box, 40% of the notes give...  [#permalink]

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18 Sep 2013, 11:31
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55% (hard)

Question Stats:

66% (02:27) correct 34% (02:32) wrong based on 135 sessions

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In a department store prize box, 40% of the notes give the winner a dreamy vacation; the other notes are blank. What is the approximate probability that 3 out of 5 people that draw the notes one after the other, and immediately return their note into the box get a dreamy vacation?

a) 0.12
b) 0.23
c) 0.35
d) 0.45
e) 0.65
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Joined: 02 Sep 2009
Posts: 65095
Re: In a department store prize box, 40% of the notes give...  [#permalink]

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18 Sep 2013, 11:39
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jlgdr wrote:
In a department store prize box, 40% of the notes give the winner a dreamy vacation; the other notes are blank. What is the approximate probability that 3 out of 5 people that draw the notes one after the other, and immediately return their note into the box get a dreamy vacation?

a) 0.12
b) 0.23
c) 0.35
d) 0.45
e) 0.65

The probability of winning is 40% = 40/100 = 2/5.
The probability of NOT winning is 60% = 3/5.

$$P(WWWNN)=\frac{5!}{3!2!}*(\frac{2}{5})^3*(\frac{3}{5})^2=\frac{144}{625}\approx{23}$$ (we multiply by $$\frac{5!}{3!2!}$$, because WWWNN scenario can occur in several ways: WWWNN, WWNWN, WNWWN, NWWWN, ... the # of such cases basically equals to the # of permutations of 5 letters WWWNN, which is $$\frac{5!}{3!2!}$$).

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Re: In a department store prize box, 40% of the notes give...  [#permalink]

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26 Aug 2015, 17:35
1
what if we do it like this?

let us assume total tickets are 20
so 8 tickets give us vacation
and 12 are blank
so 3 have to select vacation ticket

so probability$$5c3 * ((\frac{8c1*8c1*8c1)}{20c1*20c1*20c1})$$
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In a department store prize box, 40% of the notes give...  [#permalink]

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26 Aug 2015, 18:39
mahakmalik wrote:
what if we do it like this?

let us assume total tickets are 20
so 8 tickets give us vacation
and 12 are blank
so 3 have to select vacation ticket

so probability$$5c3 * ((\frac{8c1*8c1*8c1)}{20c1*20c1*20c1})$$

Hello mahakmalik,

How it was solved above is from a formula specifically for the probability when x will occur and when not x will occur, the Bernoulli formula. The initial part of your expression selects three out of five people correctly. However, the second part of the expression is essentially the same thing except for the probability for not x, which is an equally important part as anything else. Thus, extend the expression with the probability of not x:

$$5c3 * ((\frac{8c1*8c1*8c1)}{20c1*20c1*20c1})*(\frac{12c1*12c1}{20c1*20c1}))$$

Then, you have a identical formula but with bigger numbers, which leads to the same answer. Hope this helps!

Kr,
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Re: In a department store prize box, 40% of the notes give...  [#permalink]

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20 Dec 2015, 17:24
ok, first, I did not how to tackle this question...
so we we need to have 3 winners out of 5. this can be done in 5C3 ways, or 10 ways.
WWWLL
LWWWL
LLWWW
etc. so 10 ways.

now, the probability of winning is 4/10, and probability of loosing is 6/10
i did not simplify, since working with 10's are much easier.

so 4/10 must be 3 times, and 6/10 twice.
(4/10)^3*(6/10)^2 * 10 (10 ways we can arrange winners and losers).
ok, so
(4/10)^3 = 64/1000
(6/10)^2 = 36/100
now, the first one can be multiplied by 10, and rewritten as 64/100
now, multiply 64/100 with 36/100. this will be equal to 2304/10000
this is aprox. 23/100 or 0.23
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Re: In a department store prize box, 40% of the notes give...  [#permalink]

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20 Dec 2015, 21:08
I just have one question regarding this,
3 out of 5 people that draw one after the other, doesnt it suggest that the 3 people should be consucutive?

like:
5 people : [Draw,Draw,Draw,Not Draw,Not Draw] or [Not Draw,Draw,Draw,Draw,Not Draw] or [Not Draw,Not Draw,Draw,Draw,Draw]?

Any people who consider this a valid doubt , please consider Kudos.
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Joined: 21 Dec 2015
Posts: 4
Re: In a department store prize box, 40% of the notes give...  [#permalink]

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22 Dec 2015, 11:43
Quote:
The probability of winning is 40% = 40/100 = 2/5.
The probability of NOT winning is 60% = 3/5.

P(WWWNN)=5!3!2!∗(25)3∗(35)2=144625≈23 (we multiply by 5!3!2!, because WWWNN scenario can occur in several ways: WWWNN, WWNWN, WNWWN, NWWWN, ... the # of such cases basically equals to the # of permutations of 5 letters WWWNN, which is 5!3!2!).

Why do we have to divide by 3!2!? I thought that simply multiply it by 5! as it is permutation.
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Re: In a department store prize box, 40% of the notes give...  [#permalink]

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22 Dec 2015, 11:47
1
zurich wrote:
Quote:
The probability of winning is 40% = 40/100 = 2/5.
The probability of NOT winning is 60% = 3/5.

P(WWWNN)=5!3!2!∗(25)3∗(35)2=144625≈23 (we multiply by 5!3!2!, because WWWNN scenario can occur in several ways: WWWNN, WWNWN, WNWWN, NWWWN, ... the # of such cases basically equals to the # of permutations of 5 letters WWWNN, which is 5!3!2!).

Why do we have to divide by 3!2!? I thought that simply multiply it by 5! as it is permutation.

because 3 winners can be arranged in different orders, and it doesn't matter how they are arranged.
it might be W1W2W3 or W2W1W3 -> note that winner is a winner, and they are all the same. because of this, we have to divide by 3! since we have 2 non-winners, those one can be arranged as well in 2! ways, since we do not distinguish between L1 and L2.
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Re: In a department store prize box, 40% of the notes give...  [#permalink]

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30 Dec 2015, 10:21
1
zurich wrote:
Quote:
The probability of winning is 40% = 40/100 = 2/5.
The probability of NOT winning is 60% = 3/5.

P(WWWNN)=5!3!2!∗(25)3∗(35)2=144625≈23 (we multiply by 5!3!2!, because WWWNN scenario can occur in several ways: WWWNN, WWNWN, WNWWN, NWWWN, ... the # of such cases basically equals to the # of permutations of 5 letters WWWNN, which is 5!3!2!).

Why do we have to divide by 3!2!? I thought that simply multiply it by 5! as it is permutation.

Hi Zurich,

In this case the order of the people selecting the notes does not matter. It can be looked at as a combination, instead of a permutation. The number of ways to select 3 winners out of 5 is given by 5C3 = $$\frac{5!}{3!2!}$$

It is mathematically the same as saying we will find the number of permutations of 5 people, but when there are 3 identical and 2 identical people in the group. In this case, we must divide the permutation of 5 people, 5!, by the number of arrangements of each of the identical elements, 3! and 2!.

This again equals $$\frac{5!}{3!2!}$$

The only difference is the way we explain it. Either we are choosing 3 out of 5 where the order doesn't matter, or we are arranging 3 identical and 2 identical items in a row. It works out to the same thing.

I hope that helps to clear it up!
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Re: In a department store prize box, 40% of the notes give...   [#permalink] 30 Dec 2015, 10:21