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A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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24 Dec 2018, 01:00
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A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll? A. 8/81 B. 4/27 C. 16/27 D. 2/9 E. 20/81
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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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04 Jul 2019, 08:48
Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 Official Solution: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. \(\frac{8}{81}\) B. \(\frac{4}{27}\) C. \(\frac{16}{27}\) D. \(\frac{2}{9}\) E. \(\frac{20}{81}\) The numbers greater than 4 on a die are 5 and 6. The probability of getting a number greater than 4 (so 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\) and the probability of getting 4 or less is therefore, \(1 \frac{1}{3}= \frac{2}{3}\) We need to find the probability that a number greater than 4 will first appear on the third or fourth roll. The probability of a number greater than 4 first appearing on third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\) The probability of a number greater than 4 first appearing on fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\) Thus, the probability that a number greater than 4 will first appear on the third or fourth roll \(= \frac{4}{27} + \frac{8}{81}=\frac{20}{81}\) Answer: E
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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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24 Dec 2018, 01:13
Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 Par of GMAT CLUB'S New Year's Quantitative Challenge Set
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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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24 Dec 2018, 02:21
Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 for third roll no >4 4/6*4/6*2/6 = 4/27 for 4th roll no>4 4/6*4/6*4/6*2/6 = 8/81 4/27+8/81 = 20/81 IMOE



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A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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28 Dec 2018, 11:32
Archit3110 wrote: Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 for third roll no >4 4/6*4/6*2/6 = 4/27 for 4th roll no>4 4/6*4/6*4/6*2/6 = 8/81 4/27+8/81 = 20/81 IMOE Can you explain why it is 2/6? I thought it would be 4/6 for each of the three rolls. thanks.



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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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06 Feb 2019, 08:57
MPhizzle wrote: Archit3110 wrote: Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 for third roll no >4 4/6*4/6*2/6 = 4/27 for 4th roll no>4 4/6*4/6*4/6*2/6 = 8/81 4/27+8/81 = 20/81 IMOE Can you explain why it is 2/6? I thought it would be 4/6 for each of the three rolls. thanks. MPhizzlequestion has asked that no >4 appears after 3rd or 4th roll so here 4/6 ; means : 1,2,3,4 digits and 2/6 refers to 5/6 so no >4 appearing on 3rd roll 4/6*4/6*2/6 and no>4 appearing after 4th roll 4/6*4/6*4/6*2/6 hope this helps..



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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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05 Jul 2019, 02:00
Bunuel wrote: Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 Official Solution: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. \(\frac{8}{81}\) B. \(\frac{4}{27}\) C. \(\frac{16}{27}\) D. \(\frac{2}{9}\) E. \(\frac{20}{81}\) The numbers greater than 4 on a die are 5 and 6. The probability of getting a number greater than 4 (so 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\) and the probability of getting 4 or less is therefore, \(1 \frac{1}{3}= \frac{2}{3}\) We need to find the probability that a number greater than 4 will first appear on the third or fourth roll. The probability of a number greater than 4 first appearing on third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\) The probability of a number greater than 4 first appearing on fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\) Thus, the probability that a number greater than 4 will first appear on the third or fourth roll \(= \frac{4}{27} + \frac{8}{81}=\frac{20}{81}\) Answer: E ? Hi Bunuel why did we not multiply the 2 probabilities by 3C2 and the second by 4C3?I am a bit confused



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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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05 Jul 2019, 02:07
Shishou wrote: Bunuel wrote: Bunuel wrote: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. 8/81
B. 4/27
C. 16/27
D. 2/9
E. 20/81 Official Solution: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears. What is the probability that a number greater than 4 will first appear on the third or fourth roll?
A. \(\frac{8}{81}\) B. \(\frac{4}{27}\) C. \(\frac{16}{27}\) D. \(\frac{2}{9}\) E. \(\frac{20}{81}\) The numbers greater than 4 on a die are 5 and 6. The probability of getting a number greater than 4 (so 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\) and the probability of getting 4 or less is therefore, \(1 \frac{1}{3}= \frac{2}{3}\) We need to find the probability that a number greater than 4 will first appear on the third or fourth roll. The probability of a number greater than 4 first appearing on third roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}* \frac{1}{3}=\frac{4}{27}\) The probability of a number greater than 4 first appearing on fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\) Thus, the probability that a number greater than 4 will first appear on the third or fourth roll \(= \frac{4}{27} + \frac{8}{81}=\frac{20}{81}\) Answer: E ? Hi Bunuel why did we not multiply the 2 probabilities by 3C2 and the second by 4C3?I am a bit confused Because in each case we are interested in that specific sequence of events. For example, we need \((4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4)\) and not say \((more \ than \ 4; \ 4 \ or \ less; \ 4 \ or \ less)\) because in that case condition that a number greater than 4 will first appear on the third roll would not be met.
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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un
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