Throw a pair of dice 60 times. What is the probability that the sum 7 occurs between 5 and 15 times? Throw a pair of dice 60 times. What is the probability that the sum 7 occurs between 5 and 15 times?
I know this is how you could calculate the probability of sum 7 occurring 5 times:
(60 choose 5)((1/6)^5)((5/6)^55)
I'm not sure how to calculate between 5 and 15 times though.
 A: Assume two fair dice. We can set up this problem as:
Experiment: Roll Two Fair Dice 60 times
Random Variable $S$: Sum of Face Values equals $S$even
Possible Values: 0   1   2   3   4   5   ...   14   15   ... 59   60
Consider next the following characteristics:
Dichotomous Outcomes: Success = 7; Failure = Not 7
Constant Probability: Using the same Fair Dice for all Rolls yields $P(7)$ = $\dfrac{6}{36}$ remains constant over all 60 Trials.
Independence: $P(7|Any Other Value)$ = $\dfrac{6}{36}$; prior results do not affect future results.
Since the random variable is the number of Success, we have a Binomial random variable.
Hence (between 5 and 15, not inclusive),
$P(5 < S < 15)$  $=\sum_{s=6}^{14}$ $\left(\dfrac{60}{s \cdot (60 - s)}\right)$ $\left(\dfrac{6}{36}\right)^s$ $\left(\dfrac{30}{36}\right)^{60-s}$
For inclusive, sum from 5 to 15.
A: Define $p$ to be the probability that at one roll of a pair of dice the sum is 7 (and notice that it matters whether the die are identical or not!)
Now you have a a repeating Bernoulli process that is a binomial random variable with $n=60$ and $p$ as above.
The probability that this random variable is between 5 and 15 can be written using the probability mass function of a binomial random variable.
A: Winning combinations are $(1,6), (2,5), (3,4), (4,3), (5,2)$, and $(6,1)$ and yes there are $36$ possible outcomes of a pair of fair dice so the probability of getting exactly $5$ sums of $7$ is $60 \choose 5$ $(\frac 1 6)^5$ $(\frac 5 6) ^ {55}$ which is about $3.1$%.  To get the probability of between $5$ and $15$ sums of $7$ in $60$ rolls, just sum up the probabilities of getting $5$ to $15$ if you meant inclusive or $6$ to $14$ if you meant strictly between.
Using WolframAlpha, I got about $94.6$% for inclusive, and about $88.4$% for strictly between.
Type this into WolframAlpha (www.wolframalpha.com):  
summation (60 choose x ) * ((1/6)^x) * ((5/6) ^ (60-x)) for x=5 to 15
summation (60 choose x ) * ((1/6)^x) * ((5/6) ^ (60-x)) for x=6 to 14
Since you are doing $60$ rolls and the probability of getting a sum of $7$ is $1/6$ for each roll, you would expect $10$ occurrences of that sum to appear with the highest probability of any possible number of occurrences.  Since your desired range of occurrences passes thru the "meaty" part of the expected occurrence curve, we should expect the total probability to be quite high.  $94.6$% is almost certain so in $60$ rolls, you can almost be assured that you will see $5,6,7,8,9,10,11,12,13,14$ or $15$ sums of $7$.
A: $$\frac{\sum\limits_{n=5}^{15}\binom{60}{n}\cdot6^n\cdot30^{60-n}}{36^{60}}\approx0.945962$$
A: It will roll 7 an average of 60*(6/36)=10 times, with a variance of 60(1/6)(5/6)=50/6, or standard deviation of 5/sqrt(3)=2.887.
Using the normal approximation, it will happen approximately 
$$normcdf((15.5-10)/2.887)-normcdf((4.5-10)/2.887)$$
times.
