MHB Rolling 5 Dice: Probability of At Least 3 Sixes

AI Thread Summary
The probability of rolling at least three sixes with five standard 6-sided dice is calculated by considering the total outcomes of 7776. The initial calculation mistakenly counted successful outcomes as 360, but this figure double-counted scenarios with four or five sixes. The correct approach involves calculating the combinations for exactly three, four, and five sixes, leading to a successful outcome of 276. An alternative method using the complement rule yields a probability of 23/648. The accurate probability reflects the complexities of combinatorial counting in dice rolls.
veronica1999
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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:
 
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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:

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