• Target Test Prep
    5-Day Free Trial
    5-day free, full-access trial TTP Quant

    Available with Beat the GMAT members only code

    MORE DETAILS
    Target Test Prep
  • Veritas Prep
    Free Veritas GMAT Class
    Experience Lesson 1 Live Free

    Available with Beat the GMAT members only code

    MORE DETAILS
    Veritas Prep
  • e-gmat Exclusive Offer
    Free Resources for GMAT Preparation
    Practice Questions, Videos & webinars

    Available with Beat the GMAT members only code

    MORE DETAILS
    e-gmat Exclusive Offer
  • PrepScholar GMAT
    5 Day FREE Trial
    Study Smarter, Not Harder

    Available with Beat the GMAT members only code

    MORE DETAILS
    PrepScholar GMAT
  • Kaplan Test Prep
    Free Practice Test & Review
    How would you score if you took the GMAT

    Available with Beat the GMAT members only code

    MORE DETAILS
    Kaplan Test Prep
  • Economist Test Prep
    Free Trial & Practice Exam
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    Economist Test Prep
  • Magoosh
    Magoosh
    Study with Magoosh GMAT prep

    Available with Beat the GMAT members only code

    MORE DETAILS
    Magoosh
  • EMPOWERgmat Slider
    1 Hour Free
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    EMPOWERgmat Slider

Probablity

This topic has 8 expert replies and 8 member replies
Goto page
  • 1,
  • 2
Next
parveen110 Senior | Next Rank: 100 Posts Default Avatar
Joined
17 Jan 2014
Posted:
91 messages
Thanked:
7 times

Probablity

Post Wed Jan 29, 2014 11:28 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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.

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    Post 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."

    GMAT assassins aren't born, they're made,
    Rich

    _________________
    Contact Rich at Rich.C@empowergmat.com

    Post 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
    = 13/18 = E

    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course

    Thanked by: parveen110
    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
    Nadia222 Senior | Next Rank: 100 Posts
    Joined
    15 Sep 2013
    Posted:
    78 messages
    Followed by:
    6 members
    Thanked:
    8 times
    Target GMAT Score:
    720
    Post Thu Jan 30, 2014 10:32 am
    Isn't 1*5/6*4/6*3/6 equal to 60/18?

    Nadia222 Senior | Next Rank: 100 Posts
    Joined
    15 Sep 2013
    Posted:
    78 messages
    Followed by:
    6 members
    Thanked:
    8 times
    Target GMAT Score:
    720
    Post 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.

    parveen110 Senior | Next Rank: 100 Posts Default Avatar
    Joined
    17 Jan 2014
    Posted:
    91 messages
    Thanked:
    7 times
    Post Thu Jan 30, 2014 11:41 pm
    Rich.C@EMPOWERgmat.com 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.

    Post 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.

    GMAT assassins aren't born, they're made,
    Rich

    _________________
    Contact Rich at Rich.C@empowergmat.com

    Thanked by: parveen110

    GMAT/MBA Expert

    Post 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!)

    Thanked by: parveen110
    Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!
    Anantjit Junior | Next Rank: 30 Posts Default Avatar
    Joined
    13 Aug 2015
    Posted:
    12 messages
    Post 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
    = 13/18 = E

    Cheers,
    Brent
    Hi Brent,
    why is any value on 1st dice is is 1 and not 1/6 ?

    Post 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
    = 13/18 = E

    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

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course

    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
    jervizeloy Junior | Next Rank: 30 Posts
    Joined
    08 Aug 2016
    Posted:
    13 messages
    Post 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
    = 13/18 = E

    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

    Post 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

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course

    Thanked by: jervizeloy
    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
    jervizeloy Junior | Next Rank: 30 Posts
    Joined
    08 Aug 2016
    Posted:
    13 messages
    Post 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.

    evs.teja Senior | Next Rank: 100 Posts
    Joined
    22 May 2012
    Posted:
    63 messages
    Thanked:
    1 times
    Post 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

    Post 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-counting/video/788

    Here's an article too:
    Does Order Matter? Combinations and Non-Combinations - https://www.gmatprepnow.com/articles/does-order-matter-combinations-and-non-combinations-%E2%80%93-part-iii

    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course

    Thanked by: evs.teja
    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

    Best Conversation Starters

    1 Vincen 139 topics
    2 lionsshare 53 topics
    3 Roland2rule 38 topics
    4 AbeNeedsAnswers 36 topics
    5 lheiannie07 31 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description Rich.C@EMPOWERgma...

    EMPOWERgmat

    122 posts
    2 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    122 posts
    3 image description GMATGuruNY

    The Princeton Review Teacher

    108 posts
    4 image description DavidG@VeritasPrep

    Veritas Prep

    105 posts
    5 image description Jay@ManhattanReview

    Manhattan Review

    97 posts
    See More Top Beat The GMAT Experts