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# M24-27

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Math Expert
Joined: 02 Sep 2009
Posts: 44351

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16 Sep 2014, 01:22
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Difficulty:

75% (hard)

Question Stats:

59% (01:37) correct 41% (02:22) wrong based on 41 sessions

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20 throws of a die produces following results:

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. $$\frac{5}{6}$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 44351

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16 Sep 2014, 01:22
2
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Expert's post
Official Solution:

20 throws of a die produces following results:

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.

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Joined: 16 Jul 2014
Posts: 35

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

20 throws of a die produces following results:

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.
Intern
Joined: 13 Jul 2016
Posts: 10

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22 Dec 2016, 21:39
Hi,
Can you please explain in more detail please?I have not understood clearly.
Math Expert
Joined: 02 Sep 2009
Posts: 44351

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

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.

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Joined: 12 Feb 2015
Posts: 41
Location: India
GPA: 3.84

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

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.

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...!!
Math Expert
Joined: 02 Sep 2009
Posts: 44351

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

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.

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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Re: M24-27   [#permalink] 02 Oct 2017, 11:30
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# M24-27

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