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

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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

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**

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## Comments

_{4}C_{3}* (1/6)^{3}^{3}^{3}_{4}C_{3}* (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