Rolling 5 Dice: Probability of At Least 3 Sixes

Click For Summary
SUMMARY

The probability of rolling at least 3 sixes with five standard 6-sided dice is calculated to be 276/7776. The total number of outcomes for rolling five dice is 7776. The correct approach involves counting the successful outcomes for exactly three, four, and five sixes, rather than double-counting. An alternative method using the complements rule yields a probability of 23/648.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with combinatorial counting methods
  • Knowledge of the complements rule in probability
  • Ability to perform calculations involving factorials and binomial coefficients
NEXT STEPS
  • Study combinatorial probability techniques
  • Learn about the complements rule in probability
  • Explore advanced probability distributions
  • Practice problems involving multiple dice rolls and outcomes
USEFUL FOR

Mathematicians, educators, students studying probability, and anyone interested in understanding dice probability calculations.

veronica1999
Messages
61
Reaction score
0
Five standard 6-sided dice are rolled. What is the probability that at least 3 of them show a six?

I am surprised my answer is wrong.

First the total outcomes are 6x6x6x6x6=7776

Successful outcomes 10x6x6=360
There are 10 ways to choose 3 from 5 and then the remaining 2 can be any number.

360/7776 is not the answer.

The answer is 276/7776.:confused:
 
Physics news on Phys.org
Re: dice problem

veronica1999 said:
Five standard 6-sided dice are rolled. What is the probability that at least 3 of them show a six?

I am surprised my answer is wrong.

First the total outcomes are 6x6x6x6x6=7776

Successful outcomes 10x6x6=360
There are 10 ways to choose 3 from 5 and then the remaining 2 can be any number.

360/7776 is not the answer.

The answer is 276/7776.:confused:
You have double-counted the ways in which four or five sixes can occur. What you should do is to count the number of ways in which exactly three sixes can occur (10x5x5), then add the number of ways in which four sixes can occur (5x5), and finally the one way in which five sixes can occur.
 
Re: dice problem

Oops...
Thanks!:D
 
Re: dice problem

I would use the complements rule:

$$P(X)=1-\frac{{5 \choose 0}5^5+{5 \choose 1}5^4+{5 \choose 2}5^3}{6^5}=\frac{23}{648}$$
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 41 ·
2
Replies
41
Views
6K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K