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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Thu Jan 20, 2011 10:50 pm

If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will
be the same?

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Source: Beat The GMAT — Problem Solving | Thu Jan 20, 2011 10:50 pm

Everest: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Mon Jan 03, 2011 4:56 pmFollowed by: 1 member; Location: Calif; Thanked: 2 times

by Everest » Thu Jan 20, 2011 11:05 pm

If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will
be the same?

probability (atleast 2) = 1 - 1/6 = 5/6

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Fri Jan 21, 2011 2:29 am

yellowho wrote:If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?

For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.

Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)

Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.

Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6

Probability that all the outcomes are different = 6!/(6^6)

Required probability = 1 - [6!/(6^6)]

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Jan 21, 2011 3:25 am

Rahul@gurome wrote:
yellowho wrote:If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?
For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.

Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)

Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.

Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6

Probability that all the outcomes are different = 6!/(6^6)

Required probability = 1 - [6!/(6^6)]

quite scientific, whereas the smallest probability here is 1/6. Two events will be the same if the probability is not 1/6 -> this is included with our condition of 'at least 2 the same'. Hence P(2 the same)=1-Probability(not all the same) or 5/6

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Fri Jan 21, 2011 8:57 pm

Is there a different to do this? The OA is 319/324. It's exactly the same answer but that expression suggests that there's a different methodology. 6!/6^6 is quite the calculation.

[quote="Rahul@gurome"][quote="yellowho"]If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?[/quote]

For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.

Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)

Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.

Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6

Probability that all the outcomes are different = 6!/(6^6)

Required probability = 1 - [6!/(6^6)][/quote]

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Fri Jan 21, 2011 11:37 pm

yellowho wrote:Is there a different to do this? The OA is 319/324. It's exactly the same answer but that expression suggests that there's a different methodology. 6!/6^6 is quite the calculation.

I believe the method is same as I explained.
The calculation is not at all complicated as it looks. Proceed as follows,

6!/6^6 = (1*2*3*4*5*6)/(6^6) = (4*5*6*6)/(6^6) = (4*5)/(6^4) = 5/(3*3*6*6) = 5/324

Hence, required probability = 1 - (6!/6^6) = 1 - 5/324 = 319/324

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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