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# For a particular peculiar pair of dice, the probabilities of rolling 1

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Math Expert
Joined: 02 Sep 2009
Posts: 58311
For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]

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26 Mar 2019, 05:47
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Difficulty:

55% (hard)

Question Stats:

56% (02:29) correct 44% (02:10) wrong based on 25 sessions

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For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5 and 6 on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 on the two dice?

(A) 4/63
(B) 1/8
(C) 8/63
(D) 1/6
(E) 2/7

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Joined: 02 Aug 2009
Posts: 7954
For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]

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26 Mar 2019, 06:10
Bunuel wrote:
For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5 and 6 on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 on the two dice?

(A) 4/63
(B) 1/8
(C) 8/63
(D) 1/6
(E) 2/7

we will take total chances as (1+2+3+4+5+6)=21
1 has a probability of $$\frac{1}{21}$$
2 has a probability of $$\frac{2}{21}$$
3 has a probability of $$\frac{3}{21}$$
4 has a probability of $$\frac{4}{21}$$
5 has a probability of $$\frac{5}{21}$$
6 has a probability of $$\frac{6}{21}$$

Sum of 7 is possible when we throw 1 and 6, or 2 and 5, or 3 and 4. Ecah of these can be in two ways 3,4 can be 3,4 or 4,3, and similarly others...
thus probability of a sum of 7 = 2*($$\frac{1}{21}*\frac{6}{21}$$+$$\frac{2}{21}*\frac{5}{21}$$+$$\frac{3}{21}*\frac{4}{21}$$)=$$\frac{2(1*6+2*5+3*4)}{21*21}=\frac{2*28}{21*21}=\frac{2*4}{3*21}=\frac{8}{63}$$

C
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Re: For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]

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26 Mar 2019, 06:15
1
Bunuel wrote:
For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5 and 6 on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 on the two dice?

(A) 4/63
(B) 1/8
(C) 8/63
(D) 1/6
(E) 2/7

1 + 2 + 3 .. + 6 = 6*7/2 = 21

Probability of rolling a 1 = 1/21
Probability of rolling a 2 = 2/21
etc

To get a sum of 7 on two dice, one can roll (1 and 6) or (2 and 5) or (3 and 4).

Probability of (1 and 6) = 1/21 * 6/21 * 2 = 12/21*21
Probability of (2 and 5) = 2/21 * 5/21 * 2 = 20/21*21
Probability of (3 and 4) = 3/21 * 4/21 * 2 = 24/21*21

(You multiply each probability by 2 because you can get (1 and 6) in two ways - 1 on first die and 6 on second or 6 on first die and 1 on second.
and so on...

Total probability = (12 + 20 + 24)/21*21 = 56/21*21 = 8/63

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Re: For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]

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26 Mar 2019, 11:52
chetan2u wrote:
Bunuel wrote:
For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5 and 6 on each die are in the ratio 1:2:3:4:5:6. What is the probability of rolling a total of 7 on the two dice?

(A) 4/63
(B) 1/8
(C) 8/63
(D) 1/6
(E) 2/7

we will take total chances as (1+2+3+4+5+6)=21
1 has a probability of $$\frac{1}{21}$$
2 has a probability of $$\frac{2}{21}$$
3 has a probability of $$\frac{3}{21}$$
4 has a probability of $$\frac{4}{21}$$
5 has a probability of $$\frac{5}{21}$$
6 has a probability of $$\frac{6}{21}$$

Sum of 7 is possible when we throw 1 and 6, or 2 and 5, or 3 and 4. Ecah of these can be in two ways 3,4 can be 3,4 or 4,3, and similarly others...
thus probability of a sum of 7 = 2*($$\frac{1}{21}*\frac{6}{21}$$+$$\frac{2}{21}*\frac{5}{21}$$+$$\frac{3}{21}*\frac{4}{21}$$)=$$\frac{2(1*6+2*5+3*4)}{21*21}=\frac{2*28}{21*21}=\frac{2*4}{3*21}=\frac{8}{21}$$

C

chetan2u ; please edit highlighted part
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Re: For a particular peculiar pair of dice, the probabilities of rolling 1   [#permalink] 26 Mar 2019, 11:52
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