SOLUTION: You roll a single die 10 times:
a. probability that at least one resut will be divisible by three.
b. probability of getting exactly 2 sixes.
Thank you in advance
Algebra.Com
Question 320800: You roll a single die 10 times:
a. probability that at least one resut will be divisible by three.
b. probability of getting exactly 2 sixes.
Thank you in advance
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
In order for the result of rolling one die to be divisible by 3, the result of rolling the die must be either a 3 or a 6. Since there are six possible results and only two of them are considered a success, the probability of success on any one trial is 
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
The probability of at least successes in trials is equal to 1 minus the probability of at most 
successes in trials, or:
\ =\ 1\ -\ P_n(\leq k\,-\,1)\ =\ \ 1\ -\ \sum_{i\,=\,0}^{k\,-\,1} \left(n\cr i\right\)\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i})
So for part a) you need:
\ =\ 1\ -\ P_{10}(\leq 0)\ =\ 1\ -\ \sum_{i\,=\,0}^{0} \left(10\cr\ 0\right\)\left(\frac{1}{3}\right)^i\left(\frac{2}{3}\right)^{10\,-\,i}\ =\ 1\ -\ \frac{10!}{0!(10)!}\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^{10})
(1)(0.017)\ =\ 0.983)
Or in the common vernacular, highly likely.
For part b) you need:
\ =\ \left(10\cr \ 2\right\)\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^{8})
And you can do your own arithmetic.
John
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