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# One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5

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Math Expert
Joined: 02 Sep 2009
Posts: 54376
One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5  [#permalink]

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24 Mar 2019, 23:00
1
3
00:00

Difficulty:

45% (medium)

Question Stats:

68% (01:57) correct 32% (01:36) wrong based on 38 sessions

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One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5, 5, 6, 6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

(A) 1/3
(B) 4/9
(C) 1/2
(D) 5/9
(E) 2/3

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Joined: 27 Sep 2018
Posts: 37
Re: One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5  [#permalink]

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24 Mar 2019, 23:17
1
Even+odd=odd

Probability, 1st die is odd and 2nd die is even = (2/3)*(2/3) = 4/9

probability 1st die is even and 2nd die is odd = (1/3)*(1/3) = 1/9

Thus, probability that sum is odd = 4/9 + 1/9 = 5/9

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Re: One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5  [#permalink]

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25 Mar 2019, 00:51
2
Bunuel wrote:
One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5, 5, 6, 6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

(A) 1/3
(B) 4/9
(C) 1/2
(D) 5/9
(E) 2/3

odd sum = even + odd integer
so
case 1 : die 1 even integer and die 2 odd integer = 2/6 * 2/6; 1/9
case 2 ; die 1 odd integer and die 2 even integer = 4/6 *4/6 ; 4/9
sum 1/9 + 4/9 ; 5/9
IMO D
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Re: One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5   [#permalink] 25 Mar 2019, 00:51
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