There are four conditions that need to be satisfied for a binomial experiment:

1) Each trial must have two outcomes.

2) Each trial must be independent.

3) All trials must be identical.

4) The probabilities of the outcomes remain constant must not change with each trial.

The only choice that satisfies all four of these conditions (and is therefore a binomial experiment) is the rock-paper-scissors scenario.

A discrete variable is a variable which can only take a countable number of values. For example, the number of times that a coin can come up heads in ten flips can only be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Thus, there are a countable number of possible outcomes (in this case 11). This is true for coin flips, but not for caterpillar length, water flow, or rates of return for stocks.

The answer is 17.5. Simply take the values for each day, add them, and divide by the total number of days to obtain the mean: 

We are required to find the mean outcome where the probability of each possible result varies--the random/weighted mean. 

First, multiply each possible outcome by the probability of that outcome occurring. 

Second, add these results together. 

First convert .

The player's three-point shooting follows a binomial distribution with  and .

On average, he thus makes  three-point shots per game.

This means he averages 12 points per game from three-point range if he tries to make 10 three-pointers per game.

Analyzing the data, there are more 6s than anything else (mode), the median is between  and     , the mean is , and the range is 

We are required to find the mean outcome where the probability of each possible result varies--the random/weighted mean.  First, multiply each possible outcome by the probability of that outcome occurring.  Second, add these results together. 

The mean for any set of random variables is additive in the sense that

The difference is also additive, so we have 

This means the mean of  is . 

The variance is additive when the random variables are independent, which they are in this case. But it's additive in the sense that for any real numbers  (even when negative), we have

.

So for this difference, we have

.

So the mean and variance are  and , respectively. In addition to that,  is normally distributed because the sum or difference of any set of independent normal random variables is also normally distributed.

If the random variables are independent, the variances are additive in the sense that 

.

So then the variance of the sum is 

.

The standard deviation is the square root of the variance, so we have

.

This probability is simple to compute:

We want to add the probability that X is greater or equal to two. This means the probability that X=2 or X=3.

Adding the necessary probabilities we arrive at the solution.