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by harsh.champ » Fri Feb 19, 2010 6:31 am
When four dice are rolled simultaneously, in how many outcomes will at least one of the dice show 6?

(A) 155
(B) 620
(C) 625
(D) 671
(E) 1296

The OA is D.
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by vijay_venky » Fri Feb 19, 2010 6:44 am
Dice showing other than 6 every time, then the number of chances are 5^4=625.
Because we need atleast one 6, 6^4-5^4=1296-625=671
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by shashank.ism » Fri Feb 19, 2010 7:10 am
harsh.champ wrote:When four dice are rolled simultaneously, in how many outcomes will at least one of the dice show 6?

(A) 155
(B) 620
(C) 625
(D) 671
(E) 1296

The OA is D.
probbaility of getting atleast one 6 = 1-(5/6)^4 = 1296-625/1296 = 671/1296
Ans D
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by ajith » Fri Feb 19, 2010 12:27 pm
harsh.champ wrote:When four dice are rolled simultaneously, in how many outcomes will at least one of the dice show 6?

(A) 155
(B) 620
(C) 625
(D) 671
(E) 1296

The OA is D.

Total no of outcomes = 6^4
No of outcomes in which 6 never features = 5^4

No of outcomes in which at least one of the dice show 6 = 6^4-5^4 =671
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by thebigkats » Thu Jan 13, 2011 4:32 pm
I came to the answer but used rather stupid way (conclusion reached after reading other solutions :) )

There are following ways that we can get at least one of the dice showing 6:
1. One dice is 6 and others are non-6
2. Two are 6 and other 2 are non-6
3. Three are 6 and the other one is non-6
4. All four are 6

For #1:
Dice 1 = 6 and Dice 2, 3, 4 are 1-5. so TOTAL combinations = 5 x 5 x 5
Dice 2 = 6 and Dice 1, 3,4 are 1-5 so TOTAL = 5x5x5

In all total for this case: 4 x (5x5x5) = 500

For #2:
Two of the dice are 6, total of 6 combinations possible (AB, AC, AD, BC, BD, CD are 6 and other 2 are 1-5)

Th other two can be 1-5 meaning that there are 5x5 = 25 ways
So total for this case = 6x25 = 150

For #3:
3 out of 4 dice are 6, total of 4 ways possible (either ABC or ABD or ACD or BCD are 6 and the other is non-6)
the other dice can be 1-5
so total combos = 4 x 5 = 20

For 4:
only one combination possible where all are 6

so TOTAL = 500 + 150 + 20 + 1 = 671
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by GMATGuruNY » Thu Jan 13, 2011 8:33 pm
harsh.champ wrote:When four dice are rolled simultaneously, in how many outcomes will at least one of the dice show 6?

(A) 155
(B) 620
(C) 625
(D) 671
(E) 1296

The OA is D.
Number of ways to get at least one 6 = Number of ways to roll a die 4 times - Number of ways to get no 6's

Number of ways to roll a die 4 times = 6^4
Number of ways to get no 6's on all 4 rolls = 5^4

To make the math easier, take advantage of the quadratic identity x^2 - y^2 = (x+y)(x-y):

Number of ways to get at least one 6 = 6^4 - 5^4 = (6^2 + 5^2)(6^2 - 5^2) = (6^2 + 5^2)(6+5)(6-5) = 61*11*1 = 671.

The correct answer is D.
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by rid » Sun Feb 27, 2011 9:34 am
"At least one of the dice show 6" translates to a formula "All outcomes minus All outcomes, excluding those having 6".
All outcomes = 6*6*6*6
All outcomes except those having 6 = 5*5*5*5
In order to save time on further multiplication and subtraction, just calculate the unit digits. For 6*6*6*6 it is 6, for 5*5*5*5 it is 5.
Now, 6-5 = 1. So, the answer will have unit digit "1". Answer is D.
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by prashant misra » Sat Nov 05, 2011 1:22 am
still not able to get mitch sir.i was trying to solve this way through the long process.6 can be any in any of the four places in four ways once,it can occur twice in six ways,thrice in four ways and all times four 6s in just one way here i went wrong .can anyone please help me out.
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by vaibhavgupta » Sat Nov 05, 2011 9:03 am
harsh.champ wrote:When four dice are rolled simultaneously, in how many outcomes will at least one of the dice show 6?

(A) 155
(B) 620
(C) 625
(D) 671
(E) 1296

The OA is D.
Total Outcomes=1296 [6*6*6*6]
with no sixes = 625 [5*5*5*5]
hence number of outcomes where at least one of the dice show 6= 1296-625=671

Hence D it is!! :)
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by jayoptimist » Wed Jan 18, 2012 11:15 pm
My Bit.

1) Events are non related.

Total No of ways we do not get a 6 when a dice is rolled = 1-(1/6) = 5/6
4 Dices are rolled, hence (5/6)^4 = 625/1296 ways in which we dont get a 6.
1-625/1296 = 671/1296.

Hence 671
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by ArunangsuSahu » Wed Jan 18, 2012 11:20 pm
Reqd =6^4-5^4-671

(D)
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by mourinhogmat1 » Fri Feb 03, 2012 7:41 pm
On a side note, I was so involved in the wording of this problem (Had my verbal hat on I guess), I was thinking how can you even answer a question which involves probability as a certainty? The word WILL means it will happen (with certainty), right?

6^4-5^4 = 671; There is 671/1296 probability at least one 6 will appear or in 671 outcomes there will be atleast one six on the four dies.
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by ronnie1985 » Sat Feb 04, 2012 2:47 am
At least one means All minus none

P (No six) = 4C4(5/6)^4 = 625/1296
No of outcomes with at least one six = 1296 - 625 = 671 (D) is answer
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by ronnie1985 » Thu Mar 29, 2012 10:17 am
At least one six means all minus none six
Now total outcomes = 6^4
No of ways in which six does not appear = 5^4
Answer = 6^4-5^4 = 671
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by Lifetron » Tue Aug 14, 2012 6:01 am
(6^4)-(5^4)=671

Easier Approach:

After getting (6^4)-(5^4), don't calculate by multiplying. We know that the units digit is 6 for 6^4 and 5 for 5^4. So, (6-5) = 1. We have only one option with units digit as 1.

Hence D !
Last edited by Lifetron on Tue Apr 09, 2013 3:08 am, edited 1 time in total.
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