Hello!

I have a question to ask on binomial distribution: *What is the least number of times a fair die should be rolled in order to ensure that: a the probability of observing at least one 6 is more than 0.9 b the probability of observing more than one 6 is more than 0.9?*

How do I use the formula Pr(X=x) = nCr(n,x) p^x (1-p)^(n-x) in my approach to this question?

'n' has to stay unknown, but I'm not sure how to format the probability being more than 0.9 and the value of x being between 1 and n.

Help would be appreciated!!

Okay, so I think for probability questions, it's often easier to have things make sense in your head BEFORE you worry about playing with equations. So, let's start by looking at a:

the probability of observing at least one 6 is more than 0.9

Okay, so we have the probability > 0.9

We want AT LEAST one 6. Here's the thing - working with "at least" is really hard, particularly when we don't know how many trials we have. But, we DO know from the law of total probability that the probability of getting "at least one 6" is the same as 1-P("at least one 6"s complement). Or, 1-P(no 6). So, this gives us:

P(at least one 6) > 0.9

1 - P(no 6) > 0.9

P(no 6) < 0.1

Okay, so what's the binomial distribution look like here? We don't know what n is... But, we do know what p is - it's the probability of either not getting a 6 (5/6) or the probability of getting a 6 (1/6). Since we're interested in NOT getting a 6 in our new equation, why don't we call the "success" not getting a 6 (5/6). So, this gives us the equation:

Alright, so what's x? Well, x is the amount of times we DON'T roll a 6, and we DON'T want to roll a 6. It can get a little confusing with the more abstract "n" in there, so let's try using some actual numbers to try and understand what's going on:

If I roll a die 5 times, how many times do I want to NOT get a 6? 5 times

If I roll a die 3 times, how many times do I want to NOT get a 6? 3 times

So, if I roll a die n times, how many times do I want to NOT get a 6? Hopefully you see the pattern - and if you don't, please ask for us to try and explain again. But yes, it would be n times

So, now our equation is:

Which can now be solved with the use of some logarithms. However, you could also make this easier on yourself, and just pick values of n until (5/6)^n is greater than 0.1 - I'm getting that n=13 by using this method on my calculator, and it only took me half a minute to do

Of course, if you're convinced to doing it the method with logarithms:

Remember that log(1)=0 AND is strictly increasing for all numbers it's defined for, so log(<1)<0, which is why I needed to swap the sign. And, since the first integer (since n HAS to be a positive integer) greater than 12.6 is 13, the answer is 13. Also, the choice of base doesn't matter - what's important is that your calculator can do the maths to that logarithm. I used base 10 because it made my life easier as I no longer own a CAS calculator or program, and so had to use excel to calculate it for me

You should be able to apply this thinking to part b, so use that part to test your understanding. If you're still struggling, let us know what you're having problems with, and we'll take it from there