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# four fair dice are rolled. what is the probability of having

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Joined: 03 Feb 2003
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four fair dice are rolled. what is the probability of having [#permalink]  24 Jun 2003, 23:26
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four fair dice are rolled. what is the probability of having exactly 2 threes?

(A) 1/36
(B) 1/72
(C) 1/6
(D) 1/3
(E) 25/216
Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay
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Kudos [?]: 0 [0], given: 0

CAn you explain why you multiply by 4C2 and not 4?

thank you
Manager
Joined: 03 Jun 2003
Posts: 84
Location: Uruguay
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Don┬┤t explain, I have already figured it out
thank you
Intern
Joined: 10 Aug 2004
Posts: 40
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Kudos [?]: 1 [0], given: 0

Looking through some old problems and I was wondering if anyone could explain how 1/5*1/5*5/6*5/6*4C2=25/316 in this problem. I have no idea.
Manager
Joined: 18 Jun 2004
Posts: 105
Location: san jose , CA
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can any one explain ?

probability of getting 3 and 3 from 2 dice = 1/36
the dice can be (1,2), (1,3), (1,4)(2,3)(2,4)3,4)
total prob . = 6/6^4
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Senior Manager
Joined: 25 Jul 2004
Posts: 274
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Kudos [?]: 7 [0], given: 0

the probability of 3 3 not3 not3 in that order is

1/6*1/6*5/6*5/6

now you want to consider all possible orderings that have 2 3's. Well, there are 4 dice, and you need 2 to be 3's, so you have 4C2 possible ways to roll 3's, times the probablity sequence above.
Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GPA: 3.6
Followers: 0

Kudos [?]: 89 [0], given: 87105

Perhaps I missing the computation but how does that equal 25/216. I understand the logic but not the math.
Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
GMAT 1: 710 Q49 V0
GMAT 2: 740 Q40 V50
GMAT 3: 700 Q48 V38
GPA: 3.6
Followers: 0

Kudos [?]: 89 [0], given: 87105

Another stupid mistake on my part. 25/1296*6=150/1296 which simplifies to 25/216.
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