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A fair die, with sides numbered from 1 to 6, is rolled repeatedly

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Bunuel
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A fair die, with sides numbered from 1 to 6, is rolled repeatedly [#permalink] New post   24 Dec 2018, 01:00
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A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs 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 from 1 to 6, is rolled repeatedly [#permalink] New post   04 Jul 2019, 08:48
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A fair die, with sides numbered from 1 to 6, is rolled repeatedly until a number greater than 4 first appears. What is the probability that this occurs on the third or fourth roll?

A. \(\frac{8}{81}\)
B. \(\frac{4}{27}\)
C. \(\frac{16}{27}\)
D. \(\frac{2}{9}\)
E. \(\frac{20}{81}\)


A die has two numbers greater than 4: 5 and 6. The probability of rolling a number greater than 4 (either 5 or 6) is \(\frac{2}{6} = \frac{1}{3}\). Consequently, the probability of rolling a 4 or less is \(1- \frac{1}{3}= \frac{2}{3}\).

We want to determine 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 the 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 the fourth roll \(= P(4 \ or \ less; \ 4 \ or \ less; \ 4 \ or \ less; \ more \ than \ 4) = \frac{2}{3}* \frac{2}{3}*\frac{2}{3}* \frac{1}{3}=\frac{8}{81}\)

Therefore, the combined probability that a number greater than 4 will first appear on either the third or fourth roll is \(\frac{4}{27} + \frac{8}{81} = \frac{20}{81}\).


Answer: E
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Re: A fair die, with sides numbered from 1 to 6, is rolled repeatedly [#permalink] New post   24 Dec 2018, 01:13
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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


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Re: A fair die, with sides numbered from 1 to 6, is rolled repeatedly [#permalink] New post   24 Dec 2018, 02:21
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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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Re: A fair die, with sides numbered from 1 to 6, is rolled repeatedly [#permalink] 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 from 1 to 6, is rolled repeatedly [#permalink] 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 from 1 to 6, is rolled repeatedly [#permalink] 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
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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 from 1 to 6, is rolled repeatedly [#permalink] New post   05 Jul 2019, 02:07
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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 from 1 to 6, is rolled repeatedly [#permalink] New post   16 Oct 2022, 01:48
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Given that 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 and We need to find What is the probability that a number greater than 4 will first appear on the third or fourth roll?

At each stage The Probability of Getting a number Greater than 4 = \(\frac{2}{6}\) (As there are two numbers greater than 4 i.e. 5 and 6 out of 1 to 6) = \(\frac{1}{3}\)

Similarly, Probability of Getting a number less than or equal to 4 = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll)

P(number greater than 4 will first appear on the third roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{1}{3}\) = \(\frac{4}{27}\)

P(number greater than 4 will first appear on the fourth roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give ≤ 4) * P(Fourth Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) *\(\frac{1}{3}\) = \(\frac{8}{81}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll) = \(\frac{4}{27}\) + \(\frac{8}{81}\) = \(\frac{(12 + 8)}{81}\) = \(\frac{20}{81}\)

So, Answer will be E
Hope it helps!

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Re: A fair die, with sides numbered from 1 to 6, is rolled repeatedly [#permalink]
16 Oct 2022, 01:48
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