[Statistical Physics] No. of dice rolls so prob of getting a 6 is >90%

[Flucky]

This question is really wearing me out, any ideas?

Homework Statement

How many times must I throw a fair dice in order to make the probability of getting at least one ‘6’ greater than 90%?
With this many throws, what is the probability of getting three ‘6’s?

It's the first part I'm struggling with.

Homework Equations

The relevant question is the one on the top of the pic attached.

Where
P = probability of outcome interested in
N = number of throws

The Attempt at a Solution

Attatched.

I'm using P(1) as the probability of getting 1 six, so n=1.
For the second part of the equation I would use P(3).

 

[dauto]

You are trying to calculate the probability of getting one 6. The question is what's the probability of getting AT LEAST one six. that includes two 6s, three 6s, etc. Try calculating the probability of getting NO 6s. That should be 10% since 90% of the time you're getting at least one 6.

 

[TSny]

You want to consider the probability of getting at least one six. That would include the probability of getting two sixes, or three sixes, ... or N sixes.

How does the probability of getting at least one six in N throws relate to the probability of getting no sixes in N throws?

edit: I've added nothing to what dauto already said.

 

[Flucky]

"At least"... such a rookie mistake missing something like that. It's unreal how much time I've just wasted.

Thanks, I've got it now.

 

[HalfLostAndSearching]

I understand that calculating probabilities can be challenging and time-consuming. However, there are some key concepts from statistical physics that can help us approach this problem.

Firstly, we can use the concept of "independence" to calculate the probability of getting at least one '6' in a series of dice rolls. This means that each roll is not affected by the previous one, and the probability of getting a '6' on any given roll is always 1/6.

Using this concept, we can calculate the probability of not getting a '6' on any given roll as 5/6. Therefore, the probability of not getting a '6' for n rolls in a row is (5/6)^n. And the probability of getting at least one '6' in n rolls is 1 - (5/6)^n.

So, to make the probability of getting at least one '6' greater than 90%, we can set up the following equation:

1 - (5/6)^n > 0.9

Solving for n, we get n > 22. Therefore, you would need to throw the dice at least 23 times to make the probability of getting at least one '6' greater than 90%.

For the second part of the question, we can use the concept of "binomial distribution" to calculate the probability of getting exactly three '6's in a series of dice rolls. The formula for this is:

P(k) = (n choose k) * p^k * (1-p)^(n-k)

Where:
n = total number of trials (in this case, the number of dice rolls)
k = number of successes (in this case, 3 '6's)
p = probability of success (in this case, 1/6)

Substituting these values, we get:

P(3) = (n choose 3) * (1/6)^3 * (5/6)^(n-3)

Since we already know that n > 22, we can calculate the probability of getting exactly three '6's for n=23, 24, 25, etc. and see which value gives us a probability greater than 90%.

I hope this helps you with your problem. Remember, as a scientist, it's important to approach problems with a logical and systematic mindset, and to use relevant concepts and equations to guide your calculations