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May 22nd, 2019, 01:09 PM  #1 
Newbie Joined: Oct 2018 From: UK Posts: 8 Thanks: 0  Prob. of Rolling a Dice 5 Times and it Landing on 1 at least 3 times
I posted a question a few months ago regarding the probability of a Coin landing on Heads 6 times or more out of a total of 10 Throws (assuming the probability is exactly 50/50). Someone was helpful enough to break down a formula which I had a hard time understanding into an easy to understand example (see below):  The probability of a Coin landing on Heads 6 times or more is: $\dfrac{1}{1024} * \left ( \dfrac{10!}{6! * 4!} + \dfrac{10!}{7! * 3!} + \dfrac{10!}{8! * 2!} + \dfrac{10!}{9! * 1!} + \dfrac{10!}{10! * 0!} \right ) = \\ \dfrac{1}{1024} * (210 + 120 + 45 + 10 + 1) \approx 37.70\%.$  I created a computer program that could work out the probability for other scenarios based on the formula above using factorials. However it did not work for anything other than coin throws. For example the formula did not work for dice throws where the probability is 1/6. Could someone please provide a breakdown of a formula (like the one above) to work out the probability of rolling a dice Five times and it landing on the number 1 Three times or more (assuming the probability is exactly 1/6. I can then extrapolate the example to work for very different scenarios. Thanks 
May 22nd, 2019, 01:40 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,770 Thanks: 700 
The formula is based on the binomial distribution. Let $n$ be the number of trials and let $p$ be the probability of success and $q=1p$ the probability of failure. Then the probability of exactly $k$ successes in $n$ trials is $\binom{n}{k}p^kq^{nk}$ In your case $n=5$ and $p=1/6$ so what you want is$\sum_{k=3}^5\binom{5}{k}(\frac{1}{6})^k(\frac{5 }{6})^{5k}$. Note $\binom{n}{k}=\frac{n!}{k!(nk)!}$ is standard notation for binomial term. 
May 22nd, 2019, 02:42 PM  #3  
Newbie Joined: Oct 2018 From: UK Posts: 8 Thanks: 0  Quote:
My maths relating to probability is pretty basic so i'm not even sure what the symbol means. I can't google it as I don't know it's name.  
May 22nd, 2019, 02:50 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342  Quote:
\dfrac{5!}{4!}\left(\dfrac 1 6\right)^4\left(\dfrac 5 6\right) + \left(\dfrac 1 6\right)^5$  
May 22nd, 2019, 03:49 PM  #5  
Newbie Joined: Oct 2018 From: UK Posts: 8 Thanks: 0  Quote:
Also do the powers of 3, 2, 4 and 5 remain the same regardless of the number of successes, trials and probability? Is it possible to provide a very similar formula that includes say 7 roles of the dice where 4 or more rolls successfully land on 1. I could then extrapolate from your previous example how the formula would work for more complex scenarios? Last edited by enigmatic; May 22nd, 2019 at 03:52 PM. Reason: clarification regarding the question  
May 22nd, 2019, 06:24 PM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342  Quote:
Also review the definition of $\dbinom{n}{k} = \dfrac{n!}{k!(nk)!}$  

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dice, landing, prob, rolling, times 
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