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# What is the probability of rolling a 1 exactly 3 times?

You roll a number cube 4 times. What is the probability of rolling a 1 exactly 3 times?

### 2 Answers by Expert Tutors

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Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
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you roll the dice 4 times
the first time you get a 1,2,3,4,5, or 6
the second time you roll the dice you get a 1,2,3,4,5, or 6 where each number is paired with one
of the first 6 numbers for 36 combinations
roll the dice a third time and the same six numbers (each one) is paired with the 36 combinations
for a total of 36*6=216 combinations
roll the dice a fourth time and these six numbers are each paired with the 216 combinations for a total of
6*216=1296 combinations
how many of these 1296 combinations are favorable ? ( meaning how many yield exactly three ones ?)
the favorable outcomes are 1,1,1, n where n=2,3,4,5,or 6 for a total of
5 favorable outcomes, or you could have...
1,1,n,1 where n=2,3,4,5,or 6 for a total of 5 more favorable outcomes, or you could have...
1,n,1,1 where n=2,3,4,5, or 6 for 5 more favorable outcomes, or you could have...
n,1,1,1 where n=2,3,4,5, or 6 for 5 more favorable outcomes for a total of 20 favorable outcomes out of a possible 1296 combinations
the probability would be 20/1296=5/324

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
5.0 5.0 (443 lesson ratings) (443)
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Oops, in my first answer I missed the part about 4 rolls.  The corrected answer is below.  Thanks to Pierce O. for catching my mistake.

On a six-sides cube, the probability of rolling a one is 1 out of 6, or 1/6.  The probability of not rolling a 6 is 5/6.  So the probability P of rolling three  ones in four attempts is (or any number 3 consecutive times) is:

P = (1/6)3*(5/6) = 5/1296

However, there are 4 ways that you can roll 3 ones out 4 rolls:

1   1   1   not1
1   1  not1    1
1  not1   1   1
not1  1   1   1

So the probability is (5/1296)*4 = 5/324

### Comments

But they rolled the die 4 times, so this doesn't take into account all four of the rolls, it only takes the case where we do 3 rolls. Shouldn't it be

P = 4C3 * (1/6)3

= 4! / ( (4 - 3)! * 3! ) * (1/6)3

= 4 * (1/6)3

= 4 / 216

= 1 / 56
Sorry, I forgot to take into account the one roll where we did not roll a 1. The answer should be

P = 4C3 * (1/6)3 * (5/6)1
= 4! / ( (4 - 3)! * 3! ) * (1/6)3 * (5/6)

= 4 * (1/6)3 * (5/6)

= 4 * 1 / 216 * 5 / 6

= 5 / 324
I'm having a bad day reading the full question today.  I missed the 4 rolls part.