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  # For a particular peculiar pair of dice, the probabilities of rolling 1
  
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  For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]
  
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  26 Mar 2019, 05:47
  00:00
  
  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*($$\frac{1}{21}*\frac{6}{21}$$+$$\frac{2}{21}*\frac{5}{21}$$+$$\frac{3}{21}*\frac{4}{21}$$)=$$\frac{2(1*6+2*5+3*4)}{21*21}=\frac{2*28}{21*21}=\frac{2*4}{3*21}=\frac{8}{63}$$
  
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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
  
  _________________
  Karishma
  Veritas Prep GMAT Instructor
  
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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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  # For a particular peculiar pair of dice, the probabilities of rolling 1
  
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explanation: The document is a GMAT question post from a forum, specifically a probability problem involving a peculiar pair of dice. The probabilities of rolling 1, 2, 3, 4, 5, and 6 on each die are given in the ratio 1:2:3:4:5:6. The question asks for the probability of rolling a total of 7 on the two dice. The post also includes a detailed solution provided by a Veritas Prep GMAT instructor.
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distilabel_metadata.content: GMAT Question of the Day - Daily to your Mailbox; hard ones only
  
   It is currently 14 Oct 2019, 06:26
  
  ### GMAT Club Daily Prep
  
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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
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  For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]
  
  ### Show Tags
  
  26 Mar 2019, 05:47
  00:00
  
  Difficulty:
  
  55% (hard)
  
  Question Stats:
  
  56% (02:29) correct 44% (02:10) wrong based on 25 sessions
  
  ### HideShow timer Statistics
  
  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
  
  _________________
  Math Expert
  Joined: 02 Aug 2009
  Posts: 7954
  For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]
  
  ### Show Tags
  
  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
  _________________
  Veritas Prep GMAT Instructor
  Joined: 16 Oct 2010
  Posts: 9700
  Location: Pune, India
  Re: For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]
  
  ### Show Tags
  
  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
  
  _________________
  Karishma
  Veritas Prep GMAT Instructor
  
  Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
  GMAT Club Legend
  Joined: 18 Aug 2017
  Posts: 4978
  Location: India
  Concentration: Sustainability, Marketing
  GPA: 4
  WE: Marketing (Energy and Utilities)
  Re: For a particular peculiar pair of dice, the probabilities of rolling 1  [#permalink]
  
  ### Show Tags
  
  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
  _________________
  If you liked my solution then please give Kudos. Kudos encourage active discussions.
  Re: For a particular peculiar pair of dice, the probabilities of rolling 1   [#permalink] 26 Mar 2019, 11:52
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  # For a particular peculiar pair of dice, the probabilities of rolling 1
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a GMAT question post from a forum, specifically a probability problem involving a peculiar pair of dice. The probabilities of rolling 1, 2, 3, 4, 5, and 6 on each die are given in the ratio 1:2:3:4:5:6. The question asks for the probability of rolling a total of 7 on the two dice. The post also includes a detailed solution provided by a Veritas Prep GMAT instructor.
  
  Label: question-and-answer