HOW TO IDENTIFY:The first thing to realise is that Binomial Distribution is used with a series of Bernoulli Trials, where a Bernoulli Trial is something that only has two outcomes. For example, flipping a coin is considered a Bernoulli Trial as there are only two possible outcomes (heads or tails). Getting an answer right in a test is a Bernoulli Trial as there are only two possible outcomes (right answer or wrong answer) that are

*completely independent* (one outcome has no effect on the next outcome). Be careful though, as there are some tricky ones to identify.

Is "Rolling a 6 on a die" a Bernoulli Trial?? Yes. There only two outcomes (rolling a 6, or not rolling a 6). Let's look at an example.

**Example:****Answer:**Let's figure out how we can determine what type of probability to use. Like I said, we have to first identify if there is a Bernoulli Trial here. Firstly, we know that the probability John hits the bullseye is 1/4. For each question you come across, ask yourself - "Are there more than two outcomes to this trial?" If there is only two, it will most likely be a Bernoulli. This trial only has two outcomes -

hitting the bullseye or

not hitting the bullseye. Therefore,

*we have a Bernoulli Trial and hence a Binomial Distribution question on our hands.*Binomial Distribution is largely done in your calculator. Searching for this example question, I looked at Exam 1s first and couldn't actually easily find one. They'll always be on CAS-enabled Exam 2.

So, there are a few aspects we need to consider in this question.

1. The probability that John hits the bullseye at least once from four throws

2. The probability that Rebecca hits the bullseye at least once from two throws

3. The ratio between these two numbers

Now that we've identified that it's a Binomial Distribution question, we can work this out. For Binomial Distribution, we always need p, n and x, where p=probability of success, n=number of trials, and x=how many successes you have. NOTE: x can be a range of values

**John:****n** = 4

**p** = 1/4

**x** = 1, 2, 3 and 4 (x>0) - We know this because we're looking for the probability that he hits *at least one bullseye.*

All this goes into our CAS (Binomial CDF). I believe this is MENU --> 5 --> 5 --> E

n = 4, p = 1/4, lower bound = 1, upper bound = 4

Therefore, Pr(x>0) = 0.6836

**Rebecca:****n** = 2

**p** = 1/2

**x** = 1 and 2 (x>0)

Into the CAS:

n = 2, p = 1/2, lower bound = 1, upper bound = 2

Therefore, Pr(x>0) = 0.75

**Ratio:**A and D are clearly not correct. Rebecca has a higher chance than John, therefore the first number must be higher than the second. That counts out C. Then you just work out what the ratios are in a decimal form.

32/27 = 1.1852

192/175 = 1.0971

Rebecca/John = 0.75/0.6836 = 1.0971

**Therefore, the answer is E (192:175)**These questions can be quite simple once you get to know them. All you need to remember is this: If the question refers to a trial that only has two outcomes, it is a Binomial Distribution question.