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# 3 questions in the tradition of Stolyar... Three fair dice

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Manager
Joined: 06 Jun 2003
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3 questions in the tradition of Stolyar... Three fair dice [#permalink]  09 Dec 2003, 07:07
3 questions in the tradition of Stolyar...

Three fair dice are rolled, what is the probability that

(1) you get 3 sixes?
(2) you DO NOT get 3 sixes?
(3) you get ATLEAST one six?
SVP
Joined: 03 Feb 2003
Posts: 1608
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Kudos [?]: 76 [0], given: 0

in the tradition... looks great!!! nice to see it!!!
Director
Joined: 28 Oct 2003
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Location: 55405
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Quote:
(1) you get 3 sixes?

(1/6)^3

Quote:
(2) you DO NOT get 3 sixes?

1-((1/6)^3)

Quote:
(3) you get ATLEAST one six?

1-((5/6)^3)
Manager
Joined: 06 Jun 2003
Posts: 59
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Let's increase the difficulty a bit, what is the probability that

you will NOT get EXACTLY 2 sixes?
Director
Joined: 28 Oct 2003
Posts: 503
Location: 55405
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Kudos [?]: 10 [0], given: 0

Quote:
Let's increase the difficulty a bit, what is the probability that

you will NOT get EXACTLY 2 sixes?

x66 (5 possibilities)
6x6 (5 possibilities)
66x (5 possibilities)

15 ways out of 6^3 to get exactly 2 sixes.

((6^3)-15)/6^3

Last edited by stoolfi on 09 Dec 2003, 16:49, edited 1 time in total.
Manager
Joined: 06 Jun 2003
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Is it 25 or 15? (3 x 5)
Director
Joined: 28 Oct 2003
Posts: 503
Location: 55405
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Kudos [?]: 10 [0], given: 0

My mistake. I edited my errant message.
CEO
Joined: 15 Aug 2003
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Re: Dice probability [#permalink]  13 Dec 2003, 01:02
bluefox420 wrote:
3 questions in the tradition of Stolyar...

Three fair dice are rolled, what is the probability that

(1) you get 3 sixes?
(2) you DO NOT get 3 sixes?
(3) you get ATLEAST one six?

Quote:
(1) you get 3 sixes?

only one way out of 216 ways => 1/216

Quote:
(2) you DO NOT get 3 sixes?

1- 1/216 =215/216

Quote:
(3) you get ATLEAST one six?
[/quote]

1- (5/6)^3 = 91/216
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 703 [0], given: 781

bluefox420 wrote:
Let's increase the difficulty a bit, what is the probability that you will NOT get EXACTLY 2 sixes?

P (Exactly two sixes) = 3 * 1/6^2 * 5/6 = 5/72

Reqd prob = 1 - 5/72 = 67/72
CEO
Joined: 15 Aug 2003
Posts: 3469
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Kudos [?]: 703 [0], given: 781

stoolfi wrote:
Quote:
Let's increase the difficulty a bit, what is the probability that

you will NOT get EXACTLY 2 sixes?

x66 (5 possibilities)
6x6 (5 possibilities)
66x (5 possibilities)

15 ways out of 6^3 to get exactly 2 sixes.

((6^3)-15)/6^3

SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

Hi stolyar

for the improved question you need to add getting 3 sixes to getting 1 six.

x66
6x6
66x
666

1/6^3 + 3 * 1/6 * 5/6 * 5/6
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

My bad. stoolfi is right on the improved question. Q is asking not getting exactly 2 sixes.
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