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TAGS: Founder  Joined: 18 Apr 2015
Posts: 6898
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Four people each roll a fair die once. [#permalink]
1
KUDOS
Expert's post 00:00

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.
[Reveal] Spoiler: OA

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

Re: Four people each roll a fair die once. [#permalink]
How plz explain
Manager  Joined: 26 Jun 2017
Posts: 104
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Kudos [?]: 41 , given: 38

Re: Four people each roll a fair die once. [#permalink]
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  , given: 6309

Re: Four people each roll a fair die once. [#permalink]
6
KUDOS
Expert's post
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 well-suited 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 up-front, 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

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

Re: Four people each roll a fair die once. [#permalink]
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 well-suited 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 up-front, 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

Thank you very much _________________

What you think, you become.

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

Re: Four people each roll a fair die once. [#permalink]
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 well-suited 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 up-front, 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

$$1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}$$ $$= 1 - \frac{5}{18} = \frac{13}{18} = 0.72$$

What do you mean by "$$=" in the last line? I am confused. Founder  Joined: 18 Apr 2015 Posts: 6898 Followers: 114 Kudos [?]: 1342 , given: 6309 Re: Four people each roll a fair die once. [#permalink] Expert's post 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}$$ = 0.72

Regards
_________________
Intern Joined: 04 Nov 2018
Posts: 44
Followers: 0

Kudos [?]: 6 , given: 8

Re: Four people each roll a fair die once. [#permalink]
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 well-suited 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 up-front, 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

How are you so good at both the GRE Quant and Verbal?
Founder  Joined: 18 Apr 2015
Posts: 6898
Followers: 114

Kudos [?]: 1342 , given: 6309

Re: Four people each roll a fair die once. [#permalink]
Expert's post
Thank you so much for your heart words

Start reading this https://greprepclub.com/forum/gre-all-y ... -8898.html

I hope soo complete the rest.

Regards
_________________
Manager Joined: 22 Feb 2018
Posts: 163
Followers: 2

Kudos [?]: 114 , given: 22

Re: Four people each roll a fair die once. [#permalink]
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
_________________ Manager Joined: 22 Jul 2018
Posts: 79
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Kudos [?]: 11  , given: 42

Re: Four people each roll a fair die once. [#permalink]
1
KUDOS
Total = 6*6*6*6

4 people selected in 4 ways = 6*5*4*3

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  , given: 1

Re: Four people each roll a fair die once. [#permalink]
1
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 , given: 6309

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