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Probablity

by parveen110 » Wed Jan 29, 2014 11:28 pm
If 4 fair dice are thrown simultaneously, what is the probability of getting atleast one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18
OA: E

This is a veritas prep question. Although the question has already been addressed by the members on the forum, i have a confusion.

My approach:

6/6 * 1/6 * 5/6 * 4/6 = 5/54.

Where is this approach flawed?

I know the correct approach by determining the probability of getting no pairs, and then subtract that probability from 1. But I would like to know as to why the above mentioned approach is wrong. Please explain. Thank you.
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by [email protected] » Thu Jan 30, 2014 1:59 am
Hi parveen110,

The calculation that you've created is the probability that the first 2 rolls are the SAME number, the third number is different from the first two and the fourth number is different from the first three. That solution does NOT work because the question doesn't specify which two rolls must create the "pair."

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by Brent@GMATPrepNow » Thu Jan 30, 2014 7:39 am
parveen110 wrote:If 4 fair dice are thrown simultaneously, what is the probability of getting at least one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18
Rich has explained why your calculations didn't work. For others who may be wondering how we solve this, here's one approach:

We want P(get at least 1 pair)
When it comes to probability questions involving "at least," it's typically best to use the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 pair) = 1 - P(not getting at least 1 pair)
What does it mean to not get at least 1 pair? It means getting zero pairs.
So, we can write: P(getting at least 1 pair) = 1 - P(getting zero pairs)

P(getting zero pairs)
P(getting zero pairs)= P(ANY value on 1st die AND different value on 2nd die AND different value on 3rd die AND different value on 4th die)
= P(ANY value on 1st die) x P(different value on 2nd die) x P(different value on 3rd die) x P(different value on 4th die)
= 1 x 5/6 x 4/6 x 3/6
= 5/18


So, P(getting at least 1 pair) = 1 - 5/18
= [spoiler]13/18 = E[/spoiler]

Cheers,
Brent
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by Nadia222 » Thu Jan 30, 2014 10:32 am
Isn't 1*5/6*4/6*3/6 equal to 60/18?
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by Nadia222 » Thu Jan 30, 2014 10:37 am
Oops I caught my mistake the multiplication comes out to 60/216 which can them be reduced to 5/18. Got It.
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by parveen110 » Thu Jan 30, 2014 11:41 pm
[email protected] wrote:Hi parveen110,

The calculation that you've created is the probability that the first 2 rolls are the SAME number, the third number is different from the first two and the fourth number is different from the first three. That solution does NOT work because the question doesn't specify which two rolls must create the "pair."

GMAT assassins aren't born, they're made,
Rich
Hi Rich, Thanks for the reply.

Where is my logic flawed if i think that lets multiply the above mentioned step by 4C2(to consider all the possible pairs).

Thereby, giving me, 4C2*(6/6)*(1/6)*(5/6)*(4/6)?

I may be sounding silly, but i am not able to figure out why not. Thanks.
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by [email protected] » Fri Jan 31, 2014 1:00 am
Hi parveen110,

Your most recent calculation tells the you probability of getting EXACTLY one pair AND two other numbers that DON'T MATCH any of the other numbers. Unfortunately, that's not what the question is asking for either, so it's not correct.

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by Matt@VeritasPrep » Mon Feb 03, 2014 3:58 pm
One takeaway from this question: the phrase "odds of at least one" is almost always a sign that you should solve the problem by doing "100% - Odds of None" rather than by finding all the possible options and adding them together. The second approach can be insanely complicated and even if you think you've found all the valid options it's easy to forget one (or to forget some valid rearrangement of one you've already thought of!)
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by Anantjit » Wed Jul 27, 2016 6:20 am
Brent@GMATPrepNow wrote:
parveen110 wrote:If 4 fair dice are thrown simultaneously, what is the probability of getting at least one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18
Rich has explained why your calculations didn't work. For others who may be wondering how we solve this, here's one approach:

We want P(get at least 1 pair)
When it comes to probability questions involving "at least," it's typically best to use the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 pair) = 1 - P(not getting at least 1 pair)
What does it mean to not get at least 1 pair? It means getting zero pairs.
So, we can write: P(getting at least 1 pair) = 1 - P(getting zero pairs)

P(getting zero pairs)
P(getting zero pairs)= P(ANY value on 1st die AND different value on 2nd die AND different value on 3rd die AND different value on 4th die)
= P(ANY value on 1st die) x P(different value on 2nd die) x P(different value on 3rd die) x P(different value on 4th die)
= 1 x 5/6 x 4/6 x 3/6
= 5/18


So, P(getting at least 1 pair) = 1 - 5/18
= [spoiler]13/18 = E[/spoiler]

Cheers,
Brent

Hi Brent,
why is any value on 1st dice is is 1 and not 1/6 ?
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by Brent@GMATPrepNow » Wed Jul 27, 2016 7:03 am
Anantjit wrote:
Brent@GMATPrepNow wrote:
parveen110 wrote:If 4 fair dice are thrown simultaneously, what is the probability of getting at least one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18
Rich has explained why your calculations didn't work. For others who may be wondering how we solve this, here's one approach:

We want P(get at least 1 pair)
When it comes to probability questions involving "at least," it's typically best to use the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 pair) = 1 - P(not getting at least 1 pair)
What does it mean to not get at least 1 pair? It means getting zero pairs.
So, we can write: P(getting at least 1 pair) = 1 - P(getting zero pairs)

P(getting zero pairs)
P(getting zero pairs)= P(ANY value on 1st die AND different value on 2nd die AND different value on 3rd die AND different value on 4th die)
= P(ANY value on 1st die) x P(different value on 2nd die) x P(different value on 3rd die) x P(different value on 4th die)
= 1 x 5/6 x 4/6 x 3/6
= 5/18


So, P(getting at least 1 pair) = 1 - 5/18
= [spoiler]13/18 = E[/spoiler]

Cheers,
Brent

Hi Brent,
why is any value on 1st dice is is 1 and not 1/6 ?
We want the probability that ANY value (1, 2, 3, 4, 5, or 6) shows up on the first die. Since we can be sure that some value (1, 2, 3, 4, 5, or 6) will show up on the first die, the probability of this event = 1

Cheers,
Brent
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by jervizeloy » Mon Aug 08, 2016 3:08 pm
Brent@GMATPrepNow wrote:
Anantjit wrote:
Brent@GMATPrepNow wrote:
parveen110 wrote:If 4 fair dice are thrown simultaneously, what is the probability of getting at least one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18
Rich has explained why your calculations didn't work. For others who may be wondering how we solve this, here's one approach:

We want P(get at least 1 pair)
When it comes to probability questions involving "at least," it's typically best to use the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 pair) = 1 - P(not getting at least 1 pair)
What does it mean to not get at least 1 pair? It means getting zero pairs.
So, we can write: P(getting at least 1 pair) = 1 - P(getting zero pairs)

P(getting zero pairs)
P(getting zero pairs)= P(ANY value on 1st die AND different value on 2nd die AND different value on 3rd die AND different value on 4th die)
= P(ANY value on 1st die) x P(different value on 2nd die) x P(different value on 3rd die) x P(different value on 4th die)
= 1 x 5/6 x 4/6 x 3/6
= 5/18


So, P(getting at least 1 pair) = 1 - 5/18
= [spoiler]13/18 = E[/spoiler]

Cheers,
Brent

Hi Brent,
why is any value on 1st dice is is 1 and not 1/6 ?
We want the probability that ANY value (1, 2, 3, 4, 5, or 6) shows up on the first die. Since we can be sure that some value (1, 2, 3, 4, 5, or 6) will show up on the first die, the probability of this event = 1

Cheers,
Brent
Hi Brent,

How about I decide to solve this problem using counting methods. I know it's longer but I just want to make sure I'm getting the concepts clear. What would you say are the total number of possible outcomes?

Thanks
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by Brent@GMATPrepNow » Mon Aug 08, 2016 3:18 pm
jervizeloy wrote:
Hi Brent,

How about I decide to solve this problem using counting methods. I know it's longer but I just want to make sure I'm getting the concepts clear. What would you say are the total number of possible outcomes?
Each die has 6 possible outcomes, so the TOTAL number of possible outcomes = (6)(6)(6)(6)


Number of outcomes in which there are NO PAIRS
First die: 6 possibilities
Second die: 5 possibilities (anything but what was rolled on 1st die)
Third die: 4 possibilities (anything but what was rolled on 1st and 2nd dice)
Forth die: 3 possibilities
TOTAL outcomes with NO PAIRS = (6)(5)(4)(3)

So, P(NO PAIRS) = (6)(5)(4)(3)/(6)(6)(6)(6) = 5/18

This means, P(at least one pair) = 1 - 5/18 = 13/18 = E
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by jervizeloy » Mon Aug 08, 2016 5:26 pm
Brent@GMATPrepNow wrote:
jervizeloy wrote:
Hi Brent,

How about I decide to solve this problem using counting methods. I know it's longer but I just want to make sure I'm getting the concepts clear. What would you say are the total number of possible outcomes?
Each die has 6 possible outcomes, so the TOTAL number of possible outcomes = (6)(6)(6)(6)


Number of outcomes in which there are NO PAIRS
First die: 6 possibilities
Second die: 5 possibilities (anything but what was rolled on 1st die)
Third die: 4 possibilities (anything but what was rolled on 1st and 2nd dice)
Forth die: 3 possibilities
TOTAL outcomes with NO PAIRS = (6)(5)(4)(3)

So, P(NO PAIRS) = (6)(5)(4)(3)/(6)(6)(6)(6) = 5/18

This means, P(at least one pair) = 1 - 5/18 = 13/18 = E
Thanks for your kind response Brent! It's much clear now.
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by evs.teja » Fri Aug 12, 2016 3:04 am
Brent@GMATPrepNow wrote:
jervizeloy wrote:
Hi Brent,

How about I decide to solve this problem using counting methods. I know it's longer but I just want to make sure I'm getting the concepts clear. What would you say are the total number of possible outcomes?
Each die has 6 possible outcomes, so the TOTAL number of possible outcomes = (6)(6)(6)(6)


Number of outcomes in which there are NO PAIRS
First die: 6 possibilities
Second die: 5 possibilities (anything but what was rolled on 1st die)
Third die: 4 possibilities (anything but what was rolled on 1st and 2nd dice)
Forth die: 3 possibilities
TOTAL outcomes with NO PAIRS = (6)(5)(4)(3)

So, P(NO PAIRS) = (6)(5)(4)(3)/(6)(6)(6)(6) = 5/18

This means, P(at least one pair) = 1 - 5/18 = 13/18 = E
Dear Brent,

I have a small doubt , which always haunts me every time I come across such a question, will be really glad if you give your insight.

Now,
we have 4 dices, so total outcome: 6^4 (this is clear)
The Required outcome: 6*5*4*3 or 6p4
since it is a permutation order matters, so I am counting same set of numbers twice (for eg series 6,6,5,4 and 5,4,6,6 are essentially same combo but in different order, yet I am considering them as different series)
Why are we counting them twice ?

My doubt
1)How do you decide when to use permutation and when combination ?

Regards
Teja
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by Brent@GMATPrepNow » Fri Aug 12, 2016 5:55 am
evs.teja wrote: Dear Brent,

I have a small doubt , which always haunts me every time I come across such a question, will be really glad if you give your insight.

Now,
we have 4 dices, so total outcome: 6^4 (this is clear)
The Required outcome: 6*5*4*3 or 6p4
since it is a permutation order matters, so I am counting same set of numbers twice (for eg series 6,6,5,4 and 5,4,6,6 are essentially same combo but in different order, yet I am considering them as different series)
Why are we counting them twice ?

My doubt
1)How do you decide when to use permutation and when combination ?

Regards
Teja
Good question.
Notice that, for the denominator (TOTAL number of outcomes), we can saying that order matters to get 6^4
So, we need to do so for the numerator.

Here's a video on this topic:
When to use Combinations - https://www.gmatprepnow.com/module/gmat ... /video/788

Here's an article too:
Does Order Matter? Combinations and Non-Combinations - https://www.gmatprepnow.com/articles/do ... 3-part-iii

Cheers,
Brent
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