December 11, 2018 December 11, 2018 09:00 PM EST 10:00 PM EST Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST. December 13, 2018 December 13, 2018 08:00 AM PST 09:00 AM PST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 18 Aug 2005
Posts: 83
Location: Italy, Rome caput mundi

Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
Updated on: 10 Nov 2013, 03:15
Question Stats:
47% (01:57) correct 53% (02:12) wrong based on 493 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
"To win the battle you need to last 15 minutes longer than your enemies"
Originally posted by Macedon on 29 Sep 2005, 07:06.
Last edited by Bunuel on 10 Nov 2013, 03:15, edited 1 time in total.
Edited the question and added the OA.




Math Expert
Joined: 02 Sep 2009
Posts: 51100

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
11 Nov 2013, 00:58
honchos wrote: Bunuel, Your Intervention Please. This is a good question. 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 \(P(at \ least \ one \ x)=1P(no \ x's)=1\frac{4}{5}*\frac{4}{5}\). Sufficient. (2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. \(P(at \ least \ one \ x)=1P(no \ x's)=\frac{61}{125}\) > \(P(no \ x's)=\frac{64}{125}\). Since \(P(no \ x's)\) for n throws also equals to \((\frac{x1}{x})^n\), then \((\frac{x1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3\) > \(x=5\) and \(n=3\) is the only integer solution. Thus Rachel thrown the die n1=2 times and x=5 > \(P(at \ least \ one \ x)=1P(no \ x's)=1\frac{4}{5}*\frac{4}{5}\). Sufficient. Answer: D.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 04 Jul 2014
Posts: 46

Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
05 Nov 2014, 13:38
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]. PLEASE GIVE KUDOS IF THIS HELPS




VP
Joined: 22 Aug 2005
Posts: 1074
Location: CA

Re: DS: Probability
[#permalink]
Show Tags
29 Sep 2005, 08:23
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???



Manager
Joined: 14 Jul 2005
Posts: 100
Location: Sofia, Bulgaria

[#permalink]
Show Tags
Updated on: 29 Sep 2005, 13:54
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 (x1)/x. Let's say we throw the die y times, so we have
P(1 or more x) = 1  ((x1)/x)^y
and 1  ((x1)/x)^(y+1) = 9/25
=> ((x1)/x)^(y+1) = 16/25
x1 and x are consecutive numbers. For any power n, there is no way that (x1/x)^n can be presented as (simplified to) anything other than ((x1)^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, 08:35.
Last edited by vasild on 29 Sep 2005, 13:54, edited 2 times in total.



Senior Manager
Joined: 29 Nov 2004
Posts: 445
Location: Chicago

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 (x1)/x. Let's say we throw the die y times, so we have
P(1 or more x) = 1  ((x1)/x)^y
and 1  ((x1)/x)^(y+1) = 9/25
=> ((x1)/x)^(y+1) = 16/25
x1 and x are consecutive numbers. For any power n, there is no way that (x1/x)^n can be presented as (simplified to) anything other than ((x1)^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?
_________________
Fear Mediocrity, Respect Ignorance



Manager
Joined: 14 Jul 2005
Posts: 100
Location: Sofia, Bulgaria

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.



Manager
Joined: 18 Aug 2005
Posts: 83
Location: Italy, Rome caput mundi

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
_________________
"To win the battle you need to last 15 minutes longer than your enemies"



Manager
Joined: 03 Aug 2005
Posts: 129

I assume that the actual question stated on (2) 61/125
1[(x1)/x]^(y+1) = 61/125
[(x1)/x]^(y+1) = 64/125 = (4/5)^3 ===> x=5 ; y=2.



Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 483
Location: India
GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49
GPA: 3.3

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
10 Nov 2013, 02:45
Bunuel, Your Intervention Please. This is a good question.
_________________
Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/howtoscore750and750imovedfrom710to189016.html



Intern
Joined: 09 Jul 2014
Posts: 30

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
15 Oct 2014, 00:53
Bunuel wrote: honchos wrote: Bunuel, Your Intervention Please. This is a good question. 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 \(P(at \ least \ one \ x)=1P(no \ x's)=1\frac{4}{5}*\frac{4}{5}\). Sufficient. (2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. \(P(at \ least \ one \ x)=1P(no \ x's)=\frac{61}{125}\) > \(P(no \ x's)=\frac{64}{125}\). Since \(P(no \ x's)\) for n throws also equals to \((\frac{x1}{x})^n\), then \((\frac{x1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3\) > \(x=5\) and \(n=3\) is the only integer solution. Thus Rachel thrown the die n1=2 times and x=5 > \(P(at \ least \ one \ x)=1P(no \ x's)=1\frac{4}{5}*\frac{4}{5}\). Sufficient. Answer: D. Did not get it confused ,, Bunuel please help to better understand this



Math Expert
Joined: 02 Sep 2009
Posts: 51100

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
15 Oct 2014, 01:49
GuptaDarsh wrote: Bunuel wrote: honchos wrote: Bunuel, Your Intervention Please. This is a good question. 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 \(P(at \ least \ one \ x)=1P(no \ x's)=1\frac{4}{5}*\frac{4}{5}\). Sufficient. (2) If Rachel had thrown the die one more time, the probability of getting at least one x would have been 61/125. \(P(at \ least \ one \ x)=1P(no \ x's)=\frac{61}{125}\) > \(P(no \ x's)=\frac{64}{125}\). Since \(P(no \ x's)\) for n throws also equals to \((\frac{x1}{x})^n\), then \((\frac{x1}{x})^n=\frac{64}{125}=(\frac{4}{5})^3\) > \(x=5\) and \(n=3\) is the only integer solution. Thus Rachel thrown the die n1=2 times and x=5 > \(P(at \ least \ one \ x)=1P(no \ x's)=1\frac{4}{5}*\frac{4}{5}\). Sufficient. Answer: D. Did not get it confused ,, Bunuel please help to better understand this 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 09 Jul 2014
Posts: 30

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
15 Oct 2014, 05:26
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)=1P(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 (x1/x)^n, then ........"



Math Expert
Joined: 02 Sep 2009
Posts: 51100

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
15 Oct 2014, 06:28



Manager
Joined: 29 Nov 2016
Posts: 53

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
24 Feb 2018, 04:27
BunuelThe 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 + (x1)/x^2 =1/5? Posted from my mobile device



Manager
Joined: 29 Nov 2016
Posts: 53

Re: Rachel is throwing a die with sides numbered consecutively
[#permalink]
Show Tags
24 Feb 2018, 08:33
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.
Please guys help!!!
Posted from my mobile device




Re: Rachel is throwing a die with sides numbered consecutively &nbs
[#permalink]
24 Feb 2018, 08:33






