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

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In a department store prize box, 40% of the notes give... [#permalink] New post 18 Sep 2013, 11:31
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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
[Reveal] Spoiler: OA
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Re: In a department store prize box, 40% of the notes give... [#permalink] New post 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 \(\frac{5!}{3!2!}\)).

Answer: B.
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Re: In a department store prize box, 40% of the notes give... [#permalink] New post 21 Nov 2014, 10:04
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Re: In a department store prize box, 40% of the notes give... [#permalink] New post 26 Aug 2015, 17:35
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] New post 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,
Mejia
In a department store prize box, 40% of the notes give...   [#permalink] 26 Aug 2015, 18:39
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