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Probability - 3 rolls of a number cube

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Probability - 3 rolls of a number cube

by gmattesttaker2 » Sun Sep 09, 2012 12:41 am
Hello,

This is from MGMAT Guide 5. P. 63. Can you please assist with this question? Thanks a lot.

Best Regards,
Sri


What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?

Answer: [spoiler]91/216[/spoiler]

I tried solving as follows:
(1/6).(5/6).(5/6) + (1/6).(1/6).(5/6) + (1/6).(1/6).(1/6)
= 31/216


The following solution in the book is clear:
P (At least one of the rolls will be a 6 ) = 1 - P(None of the rolls will be a 6)
= 1 - (5/6).(5/6).(5/6)
= 91/216

However, I was just wondering why the above is yielding 31/216
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by neelgandham » Sun Sep 09, 2012 1:33 am
Let's focus on: P(one roll is a 6)
You wrote that this probability = (1/6)(5/6)(5/6)
This is not correct. This solution considers only the outcome of {6, not 6, not 6}
You also need to consider the outcomes {not 6, 6, not 6} and {not 6, not 6, 6}

So, P(one roll is a 6) = (1/6)(5/6)(5/6) + (5/6)(1/6)(5/6) + (5/6)(5/6)(1/6)
= 75/216

Similarly, for P(two rolls are 6), your solution of (1/6)(1/6)(5/6) considers only one possible outcome {6, 6, not 6}. You also need to consider {6, not 6, 6} and {not 6, 6, 6}

So, P(two rolls are 6) = (1/6)(1/6)(5/6) + (1/6)(5/6)(1/6) + (5/6)(1/6)(1/6)
= 15/216

Your solution for P(all 3 rolls are 6) is perfect.
P(all 3 rolls are 6) = (1/6)(1/6)(1/6)
= 1/216

So, put everything together to get:
P(at least one 6) = 75/216 + 15/216 + 1/216
= 91/216

Copied/pasted from a solution posted by Brent here - https://www.beatthegmat.com/probability- ... tml#495415
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by Brent@GMATPrepNow » Sun Sep 09, 2012 6:32 am
I should point out that the solution above (from an earlier post of mine) is meant to show the approach we'd use if we didn't use the complement: P(At least one 6 ) = 1 - P(No 6s)

Using the complement is much easier.
P(At least one 6 ) = 1 - P(No 6s)

P(No 6s) = P(non-6 on 1st roll AND non-6 on 2nd roll AND non-6 on 3rd roll)
= P(non-6 on 1st roll) X P(non-6 on 2nd roll) X P(non-6 on 3rd roll)
= (5/6)(5/6)(5/6)
= 125/216

So, P(At least one 6 ) = 1 - P(No 6s)
= 1 - 125/216
= 91/216

Cheers,
Brent
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by gmattesttaker2 » Sun Sep 09, 2012 3:07 pm
neelgandham wrote:Let's focus on: P(one roll is a 6)
You wrote that this probability = (1/6)(5/6)(5/6)
This is not correct. This solution considers only the outcome of {6, not 6, not 6}
You also need to consider the outcomes {not 6, 6, not 6} and {not 6, not 6, 6}

So, P(one roll is a 6) = (1/6)(5/6)(5/6) + (5/6)(1/6)(5/6) + (5/6)(5/6)(1/6)
= 75/216

Similarly, for P(two rolls are 6), your solution of (1/6)(1/6)(5/6) considers only one possible outcome {6, 6, not 6}. You also need to consider {6, not 6, 6} and {not 6, 6, 6}

So, P(two rolls are 6) = (1/6)(1/6)(5/6) + (1/6)(5/6)(1/6) + (5/6)(1/6)(1/6)
= 15/216

Your solution for P(all 3 rolls are 6) is perfect.
P(all 3 rolls are 6) = (1/6)(1/6)(1/6)
= 1/216

So, put everything together to get:
P(at least one 6) = 75/216 + 15/216 + 1/216
= 91/216

Copied/pasted from a solution posted by Brent here - https://www.beatthegmat.com/probability- ... tml#495415
Hello Anil,

Hope all is well. Thanks a lot for the solution. It is clear now. Thanks again for your help.

Best Regards,
Sri
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by gmattesttaker2 » Sun Sep 09, 2012 3:09 pm
Brent@GMATPrepNow wrote:I should point out that the solution above (from an earlier post of mine) is meant to show the approach we'd use if we didn't use the complement: P(At least one 6 ) = 1 - P(No 6s)

Using the complement is much easier.
P(At least one 6 ) = 1 - P(No 6s)

P(No 6s) = P(non-6 on 1st roll AND non-6 on 2nd roll AND non-6 on 3rd roll)
= P(non-6 on 1st roll) X P(non-6 on 2nd roll) X P(non-6 on 3rd roll)
= (5/6)(5/6)(5/6)
= 125/216

So, P(At least one 6 ) = 1 - P(No 6s)
= 1 - 125/216
= 91/216

Cheers,
Brent

Hello Brent,

Hope all is well. Thanks a lot for both the detailed solutions and for the excellent explanations (as always!).

Best Regards,
Sri
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