Author 
Message 
TAGS:


Founder
Joined: 18 Apr 2015
Posts: 6898
Followers: 114
Kudos [?]:
1342
[1]
, given: 6309

Four people each roll a fair die once. [#permalink]
15 Aug 2017, 01:33
1
This post received KUDOS
Question Stats:
41% (01:12) correct
58% (01:37) wrong based on 77 sessions
Four people each roll a fair die once.
Quantity A 
Quantity B 
The probability that at least two people will roll the same number 
70% 
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
_________________
Get the 2 FREE GREPrepclub Tests




Intern
Joined: 14 Mar 2017
Posts: 2
Followers: 0
Kudos [?]:
0
[0], given: 25

Re: Four people each roll a fair die once. [#permalink]
20 Aug 2017, 06:35
How plz explain



Manager
Joined: 26 Jun 2017
Posts: 104
Followers: 0
Kudos [?]:
41
[0], given: 38

Re: Four people each roll a fair die once. [#permalink]
21 Aug 2017, 05:32
Pls explain the answer. I think most of the people chose the answer B.
_________________
What you think, you become.



Founder
Joined: 18 Apr 2015
Posts: 6898
Followers: 114
Kudos [?]:
1342
[6]
, given: 6309

Re: Four people each roll a fair die once. [#permalink]
21 Aug 2017, 06:07
6
This post received KUDOS
When a question like this is really fuzzy and you do not know from where.....think smart. Now, we do know that in the first blank the question is The probability that at least two people will roll the same number. From this problem is wellsuited to the "1  x” shortcut. You can calculate the probability of each of the four rolls resulting in a different number each time. In this scenario is way much easier to calculate the probability that you DO NO want and NOT those you wanna. Do not attack the problem upfront, but using the other way around. The first roll is assigned a probability of l, since the first number that comes up will not be the same as that of any previous roll (because there has been no previous roll). In the next roll, the first number that came up must be excluded, so that there are 5 allowable outcomes. Likewise, the third roll will have 4 allowable outcomes, and the fourth roll will have 3 allowable outcomes. The overall probability of all four numbers being distinct is therefore equal to \(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}= 1  \frac{5}{18} = \frac{13}{18}\) = 0.72A is the answer
_________________
Get the 2 FREE GREPrepclub Tests



Manager
Joined: 26 Jun 2017
Posts: 104
Followers: 0
Kudos [?]:
41
[0], given: 38

Re: Four people each roll a fair die once. [#permalink]
21 Aug 2017, 09:12
Carcass wrote: When a question like this is really fuzzy and you do not know from where.....think smart.
Now, we do know that in the first blank the question is
The probability that at least two people will roll the same number.
From this problem is wellsuited to the "1  x” shortcut. You can calculate the probability of each of the four rolls resulting in a different number each time. In this scenario is way much easier to calculate the probability that you DO NO want and NOT those you wanna. Do not attack the problem upfront, but using the other way around.
The first roll is assigned a probability of l, since the first number that comes up will not be the same as that of any previous roll (because there has been no previous roll). In the next roll, the first number that came up must be excluded, so that there are 5 allowable outcomes. Likewise, the third roll will have 4 allowable outcomes, and the fourth roll will have 3 allowable outcomes. The overall probability of all four numbers being distinct is therefore equal to
\(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}= 1  \frac{5}{18} = \frac{13}{18}\) = 0.72
A is the answer Thank you very much
_________________
What you think, you become.



Director
Joined: 09 Nov 2018
Posts: 508
Followers: 0
Kudos [?]:
27
[0], given: 1

Re: Four people each roll a fair die once. [#permalink]
13 Nov 2018, 19:15
Carcass wrote: When a question like this is really fuzzy and you do not know from where.....think smart.
Now, we do know that in the first blank the question is
The probability that at least two people will roll the same number.
From this problem is wellsuited to the "1  x” shortcut. You can calculate the probability of each of the four rolls resulting in a different number each time. In this scenario is way much easier to calculate the probability that you DO NO want and NOT those you wanna. Do not attack the problem upfront, but using the other way around.
The first roll is assigned a probability of l, since the first number that comes up will not be the same as that of any previous roll (because there has been no previous roll). In the next roll, the first number that came up must be excluded, so that there are 5 allowable outcomes. Likewise, the third roll will have 4 allowable outcomes, and the fourth roll will have 3 allowable outcomes. The overall probability of all four numbers being distinct is therefore equal to
\(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}\) \(= 1  \frac{5}{18} = \frac{13}{18} = 0.72\)
A is the answer What do you mean by "\(=" in the last line? I am confused.



Founder
Joined: 18 Apr 2015
Posts: 6898
Followers: 114
Kudos [?]:
1342
[0], given: 6309

Re: Four people each roll a fair die once. [#permalink]
15 Nov 2018, 12:36
Is the Latex code that sometimes does not show properly Actually is this \(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}= 1  \frac{5}{18} = \frac{13}{18}\) = 0.72Regards
_________________
Get the 2 FREE GREPrepclub Tests



Intern
Joined: 04 Nov 2018
Posts: 44
Followers: 0
Kudos [?]:
6
[0], given: 8

Re: Four people each roll a fair die once. [#permalink]
15 Nov 2018, 12:56
Carcass wrote: When a question like this is really fuzzy and you do not know from where.....think smart.
Now, we do know that in the first blank the question is
The probability that at least two people will roll the same number.
From this problem is wellsuited to the "1  x” shortcut. You can calculate the probability of each of the four rolls resulting in a different number each time. In this scenario is way much easier to calculate the probability that you DO NO want and NOT those you wanna. Do not attack the problem upfront, but using the other way around.
The first roll is assigned a probability of l, since the first number that comes up will not be the same as that of any previous roll (because there has been no previous roll). In the next roll, the first number that came up must be excluded, so that there are 5 allowable outcomes. Likewise, the third roll will have 4 allowable outcomes, and the fourth roll will have 3 allowable outcomes. The overall probability of all four numbers being distinct is therefore equal to
\(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}= 1  \frac{5}{18} = \frac{13}{18}\) = 0.72
A is the answer Thank you very much for your detailed answer? How are you so good at both the GRE Quant and Verbal?



Founder
Joined: 18 Apr 2015
Posts: 6898
Followers: 114
Kudos [?]:
1342
[0], given: 6309

Re: Four people each roll a fair die once. [#permalink]
15 Nov 2018, 13:06
Thank you so much for your heart words Start reading this https://greprepclub.com/forum/greally ... 8898.htmlI hope soo complete the rest. Regards
_________________
Get the 2 FREE GREPrepclub Tests



Manager
Joined: 22 Feb 2018
Posts: 163
Followers: 2
Kudos [?]:
114
[0], given: 22

Re: Four people each roll a fair die once. [#permalink]
25 Nov 2018, 14:20
Answer: B
Four people each roll a fair die once. A. The probability that at least two people will roll the same number B. 70% When it says at least two people it means two, three or four people out of four. We know that P(x>=2) = P(2) + P(3) + P(4) P(a) = a people roll the same number We know P(x) = 1  P(x’) P(x>=2) = 1  P(1) P(1) = a person roll the same number or/ nobody will roll the same number, it means if the first one roll 1, second can roll between 2 and 6 and will have 5 choices not 6, the third person will have 4 choices and so on. P(1) = 6/6 * 5/6* 4/6 * 3/6 = 5/6* 1/3 = 5/18 P(x>=2) = 1  P(1) = 1  5/18 = 13/18 = 72% which is more than B
_________________
Follow your heart



Manager
Joined: 22 Jul 2018
Posts: 79
Followers: 0
Kudos [?]:
11
[1]
, given: 42

Re: Four people each roll a fair die once. [#permalink]
18 Dec 2018, 07:37
1
This post received KUDOS
Total = 6*6*6*6
4 people selected in 4 ways = 6*5*4*3
After solving answer is 5/18
The question is asked at least = 1 5/18 =0.72 which is greater than B



Intern
Joined: 27 Jan 2019
Posts: 19
Followers: 0
Kudos [?]:
18
[1]
, given: 1

Re: Four people each roll a fair die once. [#permalink]
20 Apr 2019, 05:31
1
This post received KUDOS
Hello @Carcass @chetan2u
Just for a different perspective i tried to solve this question case by case but not getting the right answer. Can u point out the mistakes in the solution? Thanks
case 1 two people throw the same number and two throw different e.g 6654
6/6*1/6*5/6*4/6 * 4!/3!2! =40/216
I am multiplying by 4! to get all the arrangements and then divide by 3! and 2! to remove the double counting.
Case 2
Three throw the same number and one throws different e.g 6664
6/6 *1/6*1/6*5/6 * 4!/3!2! =10/216
Case 3 two throw the same number and the other two throw the same but different number e.g 6655
6/6 * 1/6* 5/6*1/6 *4!/2! 2! 2! = 15/216 2! each for removing the duplicates e.g 66 and 55 and 2! more for removing the double counting of other arrangements
Case 4 All throw the same number
6/6* 1/6*1/6*1/6 =1/216
adding all cases gives 66/216 which is way off 0.72. Clearly I am missing something.



Founder
Joined: 18 Apr 2015
Posts: 6898
Followers: 114
Kudos [?]:
1342
[0], given: 6309

Re: Four people each roll a fair die once. [#permalink]
20 Apr 2019, 09:59
Hi, honestly I get lost in your approach. Sorry. However, from what I see is a process prone to errors. The easy approach is what I showed above. Kudos to you because is always worth to think the same question through the lens of a different approach. Nonetheless, The approach must be the simplest and fastest. Not cumbersome. regards
_________________
Get the 2 FREE GREPrepclub Tests




Re: Four people each roll a fair die once.
[#permalink]
20 Apr 2019, 09:59





