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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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A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un  [#permalink]

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New post 24 Dec 2018, 01:00
1
4
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A
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D
E

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  35% (medium)

Question Stats:

76% (02:22) correct 24% (02:31) wrong based on 160 sessions

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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un  [#permalink]

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New post 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  [#permalink]

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New post 24 Dec 2018, 01:13
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Re: A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled un  [#permalink]

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New post 24 Dec 2018, 02:21
2
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  [#permalink]

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New post 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  [#permalink]

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New post 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.


MPhizzle

question 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  [#permalink]

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New post 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  [#permalink]

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New post 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   [#permalink] 05 Jul 2019, 02:07
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