There are many situations where it would be interesting to determine the probability of an event occurring so many times after so many trials. For example: If I drive down the road and encounter $8$ stoplights along the way, what are the chances I make it through $5$ green lights exactly? What are the chances I have to stop at least $6$ times? Or, during a multiple choice test, each with 4 answer choices, what is the probability I answer exactly $6$ out of $10$ correctly simply by guessing?

All of these situations, and many others, follow what is called a binomial distribution only when $3$ conditions are met:

  1. The trials are independent of one another. The probability of one event occurring does not affect the probability of the next event.
  2. The probability of an event is the same on each trial.
  3. There are only two choices for each event, namely a success or a failure.

Example 1: Pier flips a fair coin $12$ times. What is the probability of exactly $5$ heads? If we were to list out the total number of ways Pier could get $5$ heads out of $12$ coin flips, it would be too exhaustive to write down. One way it could happen is $\rm \{HTHHTTHTHTTT\}$. Imagine writing down all the other ways it could happen! Luckily, from previous lessons, we recognize this as a combination problem of ${^n_r}\rm C={_5^{12}}C=\left(\dfrac{12}{5}\right)=792$. Now, from those $792$ possibilities, consider just one of them. We need to flip a head (a success) $5$ times and we need to flip a tail (a failure) $7$ times. The probability of a success is $0.5$ and we need it to happen $5$ times. The probability of a failure is $0.5$ and we need it to happen $7$ times. Therefore, the probability of $5$ heads (success) for one trial is $(0.5)^5(0.5)^7$. However this could happen $792$ different ways. Thus the total probability is $\left(\dfrac{12}{5}\right)(0.5)^5(0.5)^7=0.193$. This leads to our next formula for the binomial distribution. If we want 𝑟 success in 𝑛 trials where the probability of a success is 𝑝 and the probability of a failure is $(1−p)$, then the chances that we observe 𝑟 successes is:

$\displaystyle\mathrm{P(X}=𝑟)=\binom{n}{r}p^r(1−p)^{n−r}$ 

It is worth noting that we can use the TI-84 CE to answer this with the built in binomial distribution function under the dist command and the binompdf function.