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Re: 20 throws of a die produces following results score ==> # [#permalink]
09 Aug 2012, 06:33

Bunuel, Why are we assuming that the die is fair? Probability of a number greater than 3.5, in above example, = {# of occurrences of 4,5 and 6}/Total number of throws.

Re: 20 throws of a die produces following results score ==> # [#permalink]
09 Aug 2012, 07:21

1

This post received KUDOS

Expert's post

voodoochild wrote:

Bunuel, Why are we assuming that the die is fair? Probability of a number greater than 3.5, in above example, = {# of occurrences of 4,5 and 6}/Total number of throws.

= 8/20 Correct? Please let me know your thoughts.

Thanks

We can assume that the die is fair, because if we don't (so if we don't know that the probability of each face is 1/6), then we won't be able to solve this question at all. You cannot extrapolate the results of 20 throws to get the probability of each face if the die is unfair.

20 throws of a die produces following results

SCORE -- NUMBER OF OCCURRENCES ---1-------------------4 ---2-------------------3 ---3-------------------5 ---4-------------------2 ---5-------------------2 ---6-------------------4

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

A. 1/6 B. 1/3 C. 1/2 D. 2/3 E. 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.

Re: 20 throws of a die produces following results [#permalink]
09 Aug 2012, 07:40

Ok. But here's a question from the GMATPrep. The question doesn't assume that the random variable X's values are equally likely. Can you please explain why is that? The calculation steps are very similar. In official problem, we just have to compute the range of valid values for 'x'.

Re: 20 throws of a die produces following results [#permalink]
09 Aug 2012, 07:57

Expert's post

voodoochild wrote:

Ok. But here's a question from the GMATPrep. The question doesn't assume that the random variable X's values are equally likely. Can you please explain why is that? The calculation steps are very similar. In official problem, we just have to compute the range of valid values for 'x'.

Thanks

This question is very different from the previous one. We have that a certain experiment produced the following results:

{1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7}. Total of 15 results.

We are asked to find the probability that |x-4|>1.5, where x is randomly chosen number from the above set. Now, in order the give inequality to hold true we must choose 1, 2, 6, or 7 (or anything but 3, 4, or 5). The probability of that is 8/15.

Re: 20 throws of a die produces following results [#permalink]
01 Oct 2014, 02:22

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