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# Rachel is throwing a die with sides numbered consecutively

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Manager
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Rachel is throwing a die with sides numbered consecutively  [#permalink]

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Updated on: 10 Nov 2013, 04:15
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95% (hard)

Question Stats:

47% (01:57) correct 53% (02:13) wrong based on 501 sessions

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Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?

(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5.
(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.

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Originally posted by Macedon on 29 Sep 2005, 08:06.
Last edited by Bunuel on 10 Nov 2013, 04:15, edited 1 time in total.
Edited the question and added the OA.
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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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11 Nov 2013, 01:58
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honchos wrote:

Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?

(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5. Notice that this statement tells us that Rachel throws the die twice. Also, we are told that $$P(x)=\frac{1}{5}$$, thus $$P(not \ x)=1-\frac{1}{5}=\frac{4}{5}$$. Therefor $$P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}$$. Sufficient.

(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. $$P(at \ least \ one \ x)=1-P(no \ x's)=\frac{61}{125}$$ --> $$P(no \ x's)=\frac{64}{125}$$. Since $$P(no \ x's)$$ for n throws also equals to $$(\frac{x-1}{x})^n$$, then $$(\frac{x-1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3$$ --> $$x=5$$ and $$n=3$$ is the only integer solution. Thus Rachel thrown the die n-1=2 times and x=5 --> $$P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}$$. Sufficient.

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Rachel is throwing a die with sides numbered consecutively  [#permalink]

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05 Nov 2014, 14:38
6
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Macedon wrote:
Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?

(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5.
(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125.

statement 1 :

Probability of getting x is 1/5 . So probability of not getting an X is 4/5. Die is throw twice.

To get at least one x
-> first throw get x & second throw not get x
-> first throw not get X & second throw get X
-> get x on both rolls.

So probability of getting at least one X is 2*[1/5*4/5] + [1/5*1/5] = 9/25 . Hence statement 1 sufficient. [We need not calculate..we can stop with knowing that calculation can be done]

Statement 2:

Rachel had thrown the die for some n times , if she had thrown it for n+1 times , the probability of getting at least one x is 61/125.

When at least once is given, we can always say 1 - P (none) = P (at least once)

1 - P (none) = 61/125

P (none) = 1 - [61/125] = 64/ 125

Each throw is an independent event and hence the probability of not getting X in one throw will be the same as the probability of not getting X in the next throw and all subsequent throws if any. So, P [none] should be a perfect square only if the same fraction is multiplied two or more times.

So P[none ] = [4/5] * [4/5] * [4/5]

i.e. Rachel does three throws and gets X in none.

Now (n+1) is three throws, so n is two throws & probability of not getting X is 4/5, so probability of getting X is 1/5. Now the problem looks similar to Statement and we will get the same answer 9/25. [But we need even consider all this, we can stop with P [none] = [4/5]^3 knowing that it can be done after that]

Hence statement alone is also sufficient.

Thus Ans -> D [Note: Do not get confused with a fact that a regular die has six faces, here consider that the die Rachel uses has only 5 faces...it becomes easy].

---------------

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##### General Discussion
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29 Sep 2005, 09:23
Macedon wrote:
Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?

(1) The probability of getting an x on either of the two times that Rachel throws the die is. 1/5

(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been. 61/25

A for me too.

(1) says dice has 5 sides.
probability of NOT gettign ANY x in two throws: 4/5 * 4/5 = 16/25
therefore, probability of getting ATLEAST one x = 1 - 16/25 = 9/25

(2) 61/25 > 1. Can we have probabilities greater than 1???
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Updated on: 29 Sep 2005, 14:54
Well, it's not A, so it has to be D. btw Macedon, I assume you meant to type 16/25 and not 61/25. Even so, one more adjustment is needed: I think the second statement should read 9/25 and not 16/25. Otherwise, I don't see how this can be solved.
-----------------
Statement 1: the probability equals
1 - (4/5)^2 = 9/25, where 4/5 is the probability not to get an x

Statement 2: similar reasoning as above. The probability to get at least 1 x equals 1-(the probability to get no x's). On every throw the probability not to get an x equals (x-1)/x. Let's say we throw the die y times, so we have

P(1 or more x) = 1 - ((x-1)/x)^y

and 1 - ((x-1)/x)^(y+1) = 9/25

=> ((x-1)/x)^(y+1) = 16/25

x-1 and x are consecutive numbers. For any power n, there is no way that (x-1/x)^n can be presented as (simplified to) anything other than ((x-1)^n)/(x^n), because the two numbers do not have any common factors. I mean that 16/25 must represent the second power of 4 divided by the second power of 5.

We can therefore say that x=5 and y=1. The statement is sufficient.

Originally posted by vasild on 29 Sep 2005, 09:35.
Last edited by vasild on 29 Sep 2005, 14:54, edited 2 times in total.
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29 Sep 2005, 10:33
vasild wrote:
Well, it's not A, so it has to be D. btw Macedon, I assume you meant to type 16/25 and not 61/25. I also think there is a mistake in the second statement, and it should read 9/25. Otherwise, I don't see how this can be solved.
-----------------
Statement 1: the probability equals
1 - (4/5)^2 = 9/25, where 4/5 is the probability not to get an x

Statement 2: similar reasoning as above. The probability to get at least 1 x equals 1-(the probability to get no x's). On every throw the probability not to get an x equals (x-1)/x. Let's say we throw the die y times, so we have

P(1 or more x) = 1 - ((x-1)/x)^y

and 1 - ((x-1)/x)^(y+1) = 9/25

=> ((x-1)/x)^(y+1) = 16/25

x-1 and x are consecutive numbers. For any power n, there is no way that (x-1/x)^n can be presented as (simplified to) anything other than ((x-1)^n)/(x^n), because the two numbers do not have any common factors. I mean that 16/25 must represent the second power of 4 divided by the second power of 5.

We can therefore say that x=5 and y=1. The statement is sufficient.

I figured B is obviously not sufficient alone because it say "one more time" and we do not know one more time from what many number of times?
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29 Sep 2005, 11:36
ranga41 wrote:
I figured B is obviously not sufficient alone because it say "one more time" and we do not know one more time from what many number of times?

Ranga, I can't come up with a better explanation than what I put up above. The validity of my solution is contingent upon a wrong stem.

If thе text of the problem is exactly as in Macedon's first post, I agree with A.
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29 Sep 2005, 12:51
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THIS IS A QUESTION FROM A CAT OF PRINCETON WEBSITE....HERE IT IS THE EXPLANATION...

Yes. The pieces of the puzzle approach is useful in solving this problem. To find the probability of getting at least one x, you need to know the number of faces on the die (x) and the number of times the die is thrown. Statement (1) gives the number of times the die is thrown and also means that x = 5. The correct answer is either A or D. Statement (2) is also sufficient. To find the probability of getting at least one x, you would find the probability of not getting an x on any of the throws and subtract that probability from 1. To find the probability of not getting an x on any of the throws would require multiplying the same fraction together some number of times. Since 125^1/3= 5, the die was tossed three times in Statement (2) and it had 5 faces. The correct answer is D.

AS I SAID I PICKED A TOO..CIAO A TUTTI
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29 Sep 2005, 23:32
I assume that the actual question stated on (2) 61/125

1-[(x-1)/x]^(y+1) = 61/125
[(x-1)/x]^(y+1) = 64/125 = (4/5)^3 ===> x=5 ; y=2.
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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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10 Nov 2013, 03:45
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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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15 Oct 2014, 01:53
Bunuel wrote:
honchos wrote:

Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?

(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5. Notice that this statement tells us that Rachel throws the die twice. Also, we are told that $$P(x)=\frac{1}{5}$$, thus $$P(not \ x)=1-\frac{1}{5}=\frac{4}{5}$$. Therefor $$P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}$$. Sufficient.

(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. $$P(at \ least \ one \ x)=1-P(no \ x's)=\frac{61}{125}$$ --> $$P(no \ x's)=\frac{64}{125}$$. Since $$P(no \ x's)$$ for n throws also equals to $$(\frac{x-1}{x})^n$$, then $$(\frac{x-1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3$$ --> $$x=5$$ and $$n=3$$ is the only integer solution. Thus Rachel thrown the die n-1=2 times and x=5 --> $$P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}$$. Sufficient.

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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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15 Oct 2014, 02:49
Bunuel wrote:
honchos wrote:

Rachel is throwing a die with sides numbered consecutively from 1 to x. What is the probability that she gets at least one x?

(1) The probability of getting an x on either of the two times that Rachel throws the die is 1/5. Notice that this statement tells us that Rachel throws the die twice. Also, we are told that $$P(x)=\frac{1}{5}$$, thus $$P(not \ x)=1-\frac{1}{5}=\frac{4}{5}$$. Therefor $$P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}$$. Sufficient.

(2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. $$P(at \ least \ one \ x)=1-P(no \ x's)=\frac{61}{125}$$ --> $$P(no \ x's)=\frac{64}{125}$$. Since $$P(no \ x's)$$ for n throws also equals to $$(\frac{x-1}{x})^n$$, then $$(\frac{x-1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3$$ --> $$x=5$$ and $$n=3$$ is the only integer solution. Thus Rachel thrown the die n-1=2 times and x=5 --> $$P(at \ least \ one \ x)=1-P(no \ x's)=1-\frac{4}{5}*\frac{4}{5}$$. Sufficient.

First of all, this is a hard question from advanced topic, so it should be attempted only when all fundamentals are clear and there is no problem whatsoever in solving easier questions. Next, you really need to be more specific when asking a question.
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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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15 Oct 2014, 06:26
Bunuel : I got the 1st statement case where given is that the P(getting an x in each throw) =1/5 So, P(no X in each throw)=1- 1/5=4/5 and P(Atleast 1x)=1-P(No x)in each throw = 1- 4/5*4/5.

But in 2nd : I am lost at the point where you explained : "Since P(no x's) for n throws also equals to (x-1/x)^n, then ........"
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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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15 Oct 2014, 07:28
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Bunuel : I got the 1st statement case where given is that the P(getting an x in each throw) =1/5 So, P(no X in each throw)=1- 1/5=4/5 and P(Atleast 1x)=1-P(No x)in each throw = 1- 4/5*4/5.

But in 2nd : I am lost at the point where you explained : "Since P(no x's) for n throws also equals to (x-1/x)^n, then ........"

The probability of getting x is 1/x --> the probability of not getting x is 1 - 1/x = (x-1)/x. The probability of not getting x when throwing n times is therefore $$(\frac{x-1}{x})^n$$.
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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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24 Feb 2018, 05:27
Bunuel

The question says that probability of getting x on either throw is 1/5.

Shouldn't this mean P(getting x at throw 1) + P( getting X at throw 2) = 1/5

I.e. 1/x + (x-1)/x^2 =1/5?

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Re: Rachel is throwing a die with sides numbered consecutively  [#permalink]

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24 Feb 2018, 09:33
This actually triggers a basic question. The probability of getting a 6 on either throw of a dice, given it's thrown twice should be 1/6 + 5/36 = 11/36 i.e slightly less than 1/3. Why are we assuming the probability of X being shown on either throw to be x=5.

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Re: Rachel is throwing a die with sides numbered consecutively   [#permalink] 24 Feb 2018, 09:33
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