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Math Expert V
Joined: 02 Sep 2009
Posts: 59725

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Difficulty:   45% (medium)

Question Stats: 68% (01:44) correct 32% (02:07) wrong based on 103 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. $$\frac{5}{6}$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 59725

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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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Intern  Joined: 16 Jul 2014
Posts: 35

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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: 9

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Hi,
Can you please explain in more detail please?I have not understood clearly.
Math Expert V
Joined: 02 Sep 2009
Posts: 59725

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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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Current Student B
Joined: 12 Feb 2015
Posts: 53
Location: India
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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)

Math Expert V
Joined: 02 Sep 2009
Posts: 59725

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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)

After one more throw the number of occurrences will be 21, not 20: 68/21 = ~3.24 < 3.35.
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Joined: 31 Jul 2017
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Location: India
Schools: Anderson '21, LBS '21

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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. $$\frac{5}{6}$$

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

Moderators: chetan2u, Bunuel  