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Question of the Week 20 (A biased dice is such that any odd number..)
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26 Oct 2018, 03:51
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eGMAT Question of the Week #20A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10? A. 1/16 B. 5/72 C. 11/144 D. 11/72 E. 11/16
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Question of the Week 20 (A biased dice is such that any odd number..)
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26 Oct 2018, 04:17
EgmatQuantExpert wrote: A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10? A. 1/16 B. 5/72 C. 11/144 D. 11/72 E. 11/16 If Probability of even Number = x then Probability of odd number = 3x But, Probability of even Number on dice + Probability of odd Number on dice = 1 i.e. x+3x = 1 i.e. x = 1/4 Sum may be 10 in following ways (4,6) (5,5) (6,4) Probability of (4,6) = Probability of 4 followed by probability of 6 = (1/3)*(1/4)(1/3)*(1/4) = 1/144 Probability of (6,4) = Probability of 6 followed by probability of 4 = (1/3)*(1/4)(1/3)*(1/4) = 1/144 Probability of (5,5) = Probability of 5 followed by probability of 5 = (1/3)*(3/4)(1/3)*(3/4) = 9/144 Probability for the entire desired outcome = (1/144)+(1/144)+(9/144) = (11/144) Answer: Option C Please follow color coding to understand respective probabilities.EgmatQuantExpert I think it would have been better to mention that the dice still has six faces 3 of which are odd as the Dice may be biased on these parameters as well
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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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29 Oct 2018, 19:18
GMATinsight Hi, I am not able to understand why you multiplied 1/3 . ex: probability of 4 = 1/4*1/3 Thanks



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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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29 Oct 2018, 19:27
HAPPYatHARVARD wrote: GMATinsight Hi, I am not able to understand why you multiplied 1/3 . ex: probability of 4 = 1/4*1/3 Thanks HAPPYatHARVARDprobability of 4 = Probability of getting even number * probability of getting 4 out of 3 even numbers {2,4,6} Now probability of getting an even number on this dice = 1/4 Probability of getting 4 out of three even numbers {2,4,6} = 1/3 i.e. Probability of getting 4 on unbiased dice = 1/4*1/3 I hope this helps
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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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29 Oct 2018, 19:38
Please elaborate why did you multiply all probabilities with 1/3? GMATinsight wrote: EgmatQuantExpert wrote: A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10? A. 1/16 B. 5/72 C. 11/144 D. 11/72 E. 11/16 If Probability of even Number = x then Probability of odd number = 3x But, Probability of even Number on dice + Probability of odd Number on dice = 1 i.e. x+3x = 1 i.e. x = 1/4 Sum may be 10 in following ways (4,6) (5,5) (6,4) Probability of (4,6) = Probability of 4 followed by probability of 6 = (1/3)*(1/4)(1/3)*(1/4) = 1/144 Probability of (6,4) = Probability of 6 followed by probability of 4 = (1/3)*(1/4)(1/3)*(1/4) = 1/144 Probability of (5,5) = Probability of 5 followed by probability of 5 = (1/3)*(3/4)(1/3)*(3/4) = 9/144 Probability for the entire desired outcome = (1/144)+(1/144)+(9/144) = (11/144) Answer: Option C Please follow color coding to understand respective probabilities.EgmatQuantExpert I think it would have been better to mention that the dice still has six faces 3 of which are odd as the Dice may be biased on these parameters as well



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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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29 Oct 2018, 19:44
gvij2017 wrote: Please elaborate why did you multiply all probabilities with 1/3? GMATinsight wrote: EgmatQuantExpert wrote: A biased dice is such that any odd number appears thrice as frequently as any even number. If the dice is rolled twice, what is the probability that the sum of the numbers thus obtained is 10? A. 1/16 B. 5/72 C. 11/144 D. 11/72 E. 11/16 If Probability of even Number = x then Probability of odd number = 3x But, Probability of even Number on dice + Probability of odd Number on dice = 1 i.e. x+3x = 1 i.e. x = 1/4 Sum may be 10 in following ways (4,6) (5,5) (6,4) Probability of (4,6) = Probability of 4 followed by probability of 6 = (1/3)*(1/4)(1/3)*(1/4) = 1/144 Probability of (6,4) = Probability of 6 followed by probability of 4 = (1/3)*(1/4)(1/3)*(1/4) = 1/144 Probability of (5,5) = Probability of 5 followed by probability of 5 = (1/3)*(3/4)(1/3)*(3/4) = 9/144 Probability for the entire desired outcome = (1/144)+(1/144)+(9/144) = (11/144) Answer: Option C Please follow color coding to understand respective probabilities.EgmatQuantExpert I think it would have been better to mention that the dice still has six faces 3 of which are odd as the Dice may be biased on these parameters as well gvij2017Rationale for using the probabilities is as follows: Probability of getting an even number on dice = 1/4 Probability of getting an Odd number on dice = 3/4 Probability of getting a specific even number e.g. 2 out of three even numbers {2,4,6} = 1/3 Probability of getting a specific odd number e.g. 3 out of three odd numbers {1,3,5} = 1/3In each case either we calculate probability of a specific even number or an odd number therefore in each case we have used 1/3 for that specific even number or odd number. I hope this helps!!!
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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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01 Nov 2018, 02:17
Solution Given:• In a biased dice, any odd number appears thrice as frequently as any even number To find:• The probability that the sum of the numbers obtained upon rolling the dice twice is 10 Approach and Working: • Let us assume that the frequency of getting any even number, {2, 4, 6} is x
o Then, the frequency of getting any odd number, {1, 3, 5} will be 3x • Implies,
o The probability of getting any even number = \(\frac{x}{(x + 3x + x + 3x + x + 3x)} = \frac{1}{12}\) o And, the probability of getting any odd number = \(\frac{3x}{(x + 3x + x + 3x + x + 3x)} = \frac{3}{12} = \frac{1}{4}\) • For the sum of two outcomes to be 10, the outcomes must be either (4, 6) or (5, 5) or (6, 4) • Therefore, P(sum = 10) = P(4, 6) + P(5, 5) + P(6, 4) = \((\frac{1}{12} * \frac{1}{12}) + (\frac{1}{4} * \frac{1}{4}) + (\frac{1}{12} * \frac{1}{12}) = \frac{11}{144}\) Hence the correct answer is Option C. Answer: C
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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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02 Nov 2018, 09:55
why not probability of 4 = Probability of getting even number * probability of getting 4 out of six numbers. GMATinsight



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Re: Question of the Week 20 (A biased dice is such that any odd number..)
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02 Nov 2018, 19:52
HAPPYatHARVARD wrote: why not probability of 4 = Probability of getting even number * probability of getting 4 out of six numbers. GMATinsight HAPPYatHARVARDOnce you have included the probability of getting even number then the total choices that you are left with only three numbers {2, 4, 6} hence the following probability can NOT be 'probability of getting 4 out of six numbers', It has to be the probability of 4 out of three even numbers.
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Re: Question of the Week 20 (A biased dice is such that any odd number..) &nbs
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