# 5.4 The binomial random variable

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When you flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trialto trial. In the case of coins, heads and tails each have the same probability of 1/2. More generally, there are situations inwhich the coin is biased, so that heads and tails have different probabilities. In the present section, we consider probabilitydistributions for which there are just two possible outcomes with fixed probability summing to one. These distributions arecalled are called binomial distributions .

## A simple example

The four possible outcomes that could occur if you flipped a coin twice are listed in [link] . Note that the four outcomes are equally likely: each has probability $1/4$ . To see this, note that the tosses of the coin are independent(neither affects the other). Hence, the probability of a head on Flip 1 and a head on Flip 2 is the product of $\mathrm{P\left[H\right]}()$ and $\mathrm{P\left[H\right]}()$ , which is $1/2\times 1/2=1/4$ . The same calculation applies to the probability of a head on Flip one and a tail on Flip 2. Each is $1/2\times 1/2=1/4$ .

Four possible outcomes
Outcome First Flip Second Flip
4 Tails Tails

The four possible outcomes can be classifid in terms of the number of heads that come up. The number could betwo (Outcome 1), one (Outcomes 2 and 3) or 0 (Outcome 4). The probabilities of these possibilities are shown in [link] and in [link] . Since two of the outcomes represent the case in which just one head appears inthe two tosses, the probability of this event is equal to $1/4+1/4=1/2$ . [link] summarizes the situation.

Probabilities of getting 0,1, or 2 heads.
0 1/4
1 1/2
2 1/4

[link] is a discrete probability distribution: It shows the probability for each of the values on theX-axis. Defining a head as a "success," [link] shows the probability of 0, 1, and 2 successes for two trials (flips) for an event that has a probability of 0.5 of being asuccess on each trial. This makes [link] an example of a binomial distribution .

## The formula for binomial probabilities

The binomial distribution consists of the probabilities of each of the possible numbers of successes on $n$ trials for independent events that each have a probability of $p()$ of occurring. For the coin flip example, $n=2$ and $\mathrm{p=}(0.5)$ . The formula for the binomial distribution is shown below: $\mathrm{P\left[x\right]}()=\frac{n!}{x!(n-x)!}p^{x}(1-p)^{(n-x)}$ where $\mathrm{P\left[x\right]}()$ is the probability of $x$ successes out of $n$ trials, $n$ is the number of trials, and $p$ is the probability of success on a given trial. Applying this to thecoin flip example, $P\left[0\right]()=\frac{2!}{0!(2-0)!}0.5^{0}(1-0.5)^{(2-0)}=\frac{2}{2}\times 1\times .25=0.25$ $P\left[1\right]()=\frac{2!}{1!(2-1)!}0.5^{1}(1-0.5)^{(2-1)}=\frac{2}{1}\times .5\times .5=0.50$ $P\left[2\right]()=\frac{2!}{2!(2-2)!}0.5^{2}(1-0.5)^{(2-2)}=\frac{2}{2}\times .25\times 1=0.25$ If you flip a coin twice, what is the probability of getting one or more heads? Since the probability of getting exactlyone head is 0.50 and the probability of getting exactly two heads is 0.25, the probability of getting one or more heads is $0.50+0.25=0.75$ .

Now suppose that the coin is biased; let's say the probability of heads is only 0.4. What is the probability of getting heads at leastonce in two tosses? We could substitute p=0.4 with x=1 and with x=2 into our general formula above; adding the results would obtain the answer 0.64.

## Cumulative probabilities

We toss a coin 12 times. What is the probability that we get from 0 to 3 heads? The answer is found by computing theprobability of exactly 0 heads, exactly 1 head, exactly 2 heads, and exactly 3 heads. The probability of getting from 0to 3 heads is then the sum of these probabilities. The probabilities are: 0.0002, 0.0029, 0.0161, and 0.0537. The sumof the probabilities is 0.073. The calculation of cumulative binomial probabilities can be quite tedious. Therefore we haveprovided a binomial calculator to make it easy to calculate these probabilities.

## Mean and standard deviation of binomial distributions

Consider a coin-tossing experiment in which you tossed a coin 12 times and recorded the number of heads. If you performedthis experiment over and over again, what would the mean number of heads be? On average, you would expect half the cointosses to come up heads. Therefore the mean number of heads would be 6. In general, the mean of a binomial distributionwith parameters $n$ (the number of trials) and $p$ (the probability of success for each trial) is: $\mu =np$ where $\mu$ is the mean of the binomial distribution. The variance of the binomial distribution is: $\sigma ^{2}=np(1-p)$ where $\sigma ^{2}$ is the variance of the binomial distribution.

Let's return to the coin tossing experiment. The coin was tossed 12 times so $n=12$ . A coin has a probability of 0.5 of coming up heads. Therefore, $p=0.5$ . The mean and standard deviation can therefore be computed as follows: $\mu =np=12\times 0.5=6$ $\sigma ^{2}=np(1-p)=12\times 0.5(1.0-0.5)=3.0$ Naturally, the standard deviation $\left(\sigma \right)$ is the square root of the variance $\left(\sigma ^{2}\right)$ .

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20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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