Probability with binomial random variables

Conditions of a binomial random variable

Remember that “bi” means two, so a binomial variable is a variable that can take on exactly two values. A coin is the most obvious example of a binomial variable because flipping the coin can only result in two values: heads or tails.

On the other hand, a $6$-sided die doesn’t produce a binomial variable, since rolling the die can result in $6$ possible outcomes, not $2$.

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The two outcomes that the binomial random variable can take do not have to be equally probable. The probability of getting heads when I flip a fair coin is a binomial random variable where the probability of “success” is $50\%$. But the probability of choosing a girl from my math class when I randomly choose one student might be a binomial random variable where the probability of “success” is $70\%$ (if $70\%$ of the students in my class are female).

Binomial random variables

In order for a variable $X$ to be a binomial random variable,

• each trial must be independent,

• each trial can be called a “success” or “failure,”

• there are a fixed number of trials, and

• the probability of success on each trial is constant.

Let’s think about $X$ as how many heads we get when we flip a coin $10$ times.

• Each trial is independent, because the result of one flip doesn’t effect the result of any other flip.

• Each trial can be called a success or failure because “heads” is a success and “tails” is a failure.

• We’re flipping the coin $10$ times, so there are a fixed number of trials.

• The probability of getting heads on the first flip is $50\%$, the probability of getting heads on the second flip is $50\%$, the probability of getting heads on the third flip is $50\%$, etc. The probability of heads in each trial is constant.

So if $X$ is how many heads we get when we flip a coin $10$ times, then we could call $X$ a binomial random variable. On the other hand, let’s think about $J$ as how many jacks we get when we pull $3$ cards from a deck without replacing the card after each pull.

• Each trial is not independent, because the result of the first pull effects the result of every pull thereafter. If we get a jack on the first pull, then the probability we get a jack on the second pull is $3/51$. But if we get something other than a jack on the first pull, then the probability we get a jack on the second pull is $4/51$.

• Each trial can be called a success or failure because a jack is a success and anything else is a failure.

• We’re pulling a card $3$ times, so there are a fixed number of trials.

• The probability of getting a jack on the first pull is $4/52$. But the probability of getting a jack on the second pull changes depending on what we got on the first pull. The probability of pulling a jack in each trial is not constant.

Since it fails two of the four conditions, $J$ cannot be called a binomial random variable.

Binomial probability

In binomial probability questions, we’re often asked to figure out the probability that we get an exact number of “successes,” assuming we perform a specific number of independent trials. The formula we’ll use for this is

$P(k\text{ successes in }n\text{ attempts})=\binom{n}{k}p^k(1-p)^{n-k}$

where the binomial coefficient

$\binom{n}{k}$

is the combination $_nC_k$, $p$ is the probability of a success, $k$ is the exact number of times we want the success, and $n$ is the total number of independent trials we’ll run.

How to calculate probabilities for binomial random variables

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Pulling marbles from a bag can be modeled by a binomial random variable

Example

Let’s say there are three marbles in a bag: $2$ are green and $1$ is red. We’re going to do $5$ trials where we pull a marble, note the color, and then replace the marble. What is the probability that we get the red marble exactly $3$ times?

First, we’ll confirm that this is a binomial random variable.

1. Since we’re replacing each marble we pull before pulling another, each pull is an independent trial.

2. We can classify a red marble as a success, and green marble as a failure.

3. There are a fixed number of trials: $5$.

4. The probability of success (getting a red marble) is constant through each trial, since we’re replacing the marbles. Since all four conditions are met, this is a binomial random variable.

We’re trying to pull a red marble exactly $3$ times in $5$ pulls. To solve this problem, we need to first figure out how many possible combinations we can do this in.

$_5C_3=\binom53=\frac{5!}{3!(5-3)!}$

$_5C_3=\binom53=\frac{5!}{3!2!}$

$_5C_3=\binom53=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot2\cdot1}$

$_5C_3=\binom53=\frac{5\cdot4}{2\cdot1}$

$_5C_3=\binom53=\frac{20}{2}$

$_5C_3=\binom53=10$

Then to find the probability that we get exactly $3$ reds on $5$ pulls, we say that $p$ is the probability of pulling a red marble, and therefore that the probability of pulling $3$ red marbles in $5$ pulls is

$P(k\text{ successes in }n\text{ attempts})=\binom{n}{k}p^k(1-p)^{n-k}$

$P(3\text{ reds in }5\text{ pulls})=(10)\left(\frac13\right)^3\left(\frac23\right)^2$

$P(3\text{ reds in }5\text{ pulls})=\frac{40}{243}\approx16.5\%$

In other words, we have approximately a $16.5\%$ chance of getting exactly $3$ red marbles when we pull a random marble from this bag $5$ times.

A coin is the most obvious example of a binomial variable because flipping the coin can only result in two values: heads or tails.

Probability distributions for binomial random variables

We can also create a probability distribution for binomial random variables. Using our example of pulling a marble $5$ times, where we have a $1/3$ probability of pulling a red marble and $2/3$ probability of pulling a green marble on each pull, we could calculate the following probabilities.

$P(0\text{ red in }5\text{ pulls})=\binom{5}{0}0.33^0 0.67^5=(1)(1)(0.67^5)\approx0.1350$

$P(1\text{ red in }5\text{ pulls})=\binom{5}{1}0.33^1 0.67^4=(5)(0.33^1)(0.67^4)\approx0.3325$

$P(2\text{ red in }5\text{ pulls})=\binom{5}{2}0.33^2 0.67^3=(10)(0.33^2)(0.67^3)\approx0.3275$

$P(3\text{ red in }5\text{ pulls})=\binom{5}{3}0.33^3 0.67^2=(10)(0.33^3)(0.67^2)\approx0.1613$

$P(4\text{ red in }5\text{ pulls})=\binom{5}{4}0.33^4 0.67^1=(5)(0.33^4)(0.67^1)\approx0.0397$

$P(5\text{ red in }5\text{ pulls})=\binom{5}{5}0.33^5 0.67^0=(1)(0.33^5)(0.67^0)\approx0.0039$

We could then plot this probability distribution to get a picture of the probability.

This probability distribution is a visual representation of the fact that we’re most likely to get $1$ or $2$ red marbles when we pull $5$ times from the bag of marbles. We’re much less likely to get $4$ or $5$ red marbles when we pull $5$ times.

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