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M24-27

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M24-27  [#permalink]

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New post 16 Sep 2014, 01:22
1
4
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

68% (01:44) correct 32% (02:07) wrong based on 103 sessions

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Re M24-27  [#permalink]

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New post 16 Sep 2014, 01:22
2
Official Solution:


20 throws of a die produces following results:

Image

What is the probability that one more throw to this series will increase the mean score?


A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)

Current mean score = \(\frac{1*4 + 2*3 + 3*5 + 4*2 + 5*2 + 6*4}{20} = \frac{67}{20} = 3.35\). Only the fall of 4, 5, or 6 will increase the mean score of the series.

Answer: C
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Re: M24-27  [#permalink]

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New post 02 Nov 2014, 18:17
1
Bunuel wrote:
Official Solution:


20 throws of a die produces following results:

Image

What is the probability that one more throw to this series will increase the mean score?


Both the scores 1 and 6 have occurrences of 4, they cancel out to an average of 3.5 with half the roll outcomes increasing the mean score.

To save time we can average scores 2-5, or we can reason that since scores 4 and 5 have a total of 4 occurrences, while score 2 only has 3 occurrences, that the average will be just a bit greater than 3, and that only rolling 4, 5, and 6 will increase the average.
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Re: M24-27  [#permalink]

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New post 22 Dec 2016, 21:39
Hi,
Can you please explain in more detail please?I have not understood clearly.
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Re: M24-27  [#permalink]

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New post 23 Dec 2016, 01:30
Bunuel wrote:
20 throws of a die produces following results:

Image

What is the probability that one more throw to this series will increase the mean score?


A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)


The average score now is \(\frac{total \ score}{# \ of \ throws}=\frac{1*4+2*3+3*5+4*2+5*2+6*4}{20}=\frac{67}{20}=3.something\).

Now, the average score will increase if we get more than the current average, so if we get 4, 5, or 6 on the next throw. The probability of that is 3/6=1/2.

Answer: C.
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M24-27  [#permalink]

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New post 02 Oct 2017, 07:31
Bunuel wrote:
Bunuel wrote:
20 throws of a die produces following results:

Image

What is the probability that one more throw to this series will increase the mean score?


A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)


The average score now is \(\frac{total \ score}{# \ of \ throws}=\frac{1*4+2*3+3*5+4*2+5*2+6*4}{20}=\frac{67}{20}=3.something\).


Now, the average score will increase if we get more than the current average, so if we get 4, 5, or 6 on the next throw. The probability of that is 3/6=1/2.

Answer: C.



Bunuel

Isn't it like whenever we throw the outcome will have the result of increasing the mean,coz whenever a number is added to the set the mean,the mean will eventually increase.Suppose in this cast if the mean is 3.something and if 1or 2 or 3 is added then eventually it will not decrease the mean.
For instance if 1 is added to the set then it will become 68/20 which is 3.4>3.35(original)

Please correct my understanding...!!
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Re: M24-27  [#permalink]

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New post 02 Oct 2017, 11:30
himanshukamra2711 wrote:
Bunuel wrote:
Bunuel wrote:
20 throws of a die produces following results:

Image

What is the probability that one more throw to this series will increase the mean score?


A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)


The average score now is \(\frac{total \ score}{# \ of \ throws}=\frac{1*4+2*3+3*5+4*2+5*2+6*4}{20}=\frac{67}{20}=3.something\).


Now, the average score will increase if we get more than the current average, so if we get 4, 5, or 6 on the next throw. The probability of that is 3/6=1/2.

Answer: C.



Bunuel

Isn't it like whenever we throw the outcome will have the result of increasing the mean,coz whenever a number is added to the set the mean,the mean will eventually increase.Suppose in this cast if the mean is 3.something and if 1or 2 or 3 is added then eventually it will not decrease the mean.
For instance if 1 is added to the set then it will become 68/20 which is 3.4>3.35(original)

Please correct my understanding...!!


After one more throw the number of occurrences will be 21, not 20: 68/21 = ~3.24 < 3.35.
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M24-27  [#permalink]

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New post 19 Mar 2019, 08:25
1
Bunuel wrote:
20 throws of a die produces following results:

Image

What is the probability that one more throw to this series will increase the mean score?


A. \(\frac{1}{6}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)



Hi Bunuel, chetan2u

My answer was 2/5 Please tell me what's wrong in my reasoning

Average =3.35

So in order to increase the mean, the person has to score 4/5/6.

Now looking at his records,we can see that he can do so 8 out of 20 times he has played this game. How can we ignore the frequency ?

So the probability has to be 8/20=2/5 .

Is there anything that i am completely mistaken by ? Please let me know..

Thanks,
Manas
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M24-27   [#permalink] 19 Mar 2019, 08:25
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