"At least" and "at most," and mean, variance, and standard deviation

“At least” and “at most” probability

We can do more than just calculate the probability of pulling exactly $3$ red marbles in $5$ total pulls.

For any binomial random variable, we can also calculate something like the probability of pulling at least $3$ red marbles, or the probability of pulling no more than $3$ marbles.

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What we want to know is that the probability of pulling at least $3$ red marbles is the probability that we pull $3$, or $4$, or $5$ red marbles, which is simply the probability of each of these, all added together.

$P(\text{at least }3\text{ reds in }5\text{ pulls})=P(3\text{ reds})+P(4\text{ reds})+P(5\text{ reds})$

$P(\text{at least }3\text{ reds in }5\text{ pulls})=\binom{5}{3}\left(\frac13\right)^3 0.67^2$

$+\binom{5}{4}\left(\frac13\right)^4 \left(\frac23\right)^1+\binom{5}{5}\left(\frac13\right)^5 \left(\frac23\right)^0$

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

$+(5)\left(\frac13\right)^4\left(\frac23\right)^1+(1)\left(\frac13\right)^5\left(\frac23\right)^0$

$P(\text{at least }3\text{ reds in }5\text{ pulls})\approx0.1646+0.0412+0.0041$

$P(\text{at least }3\text{ reds in }5\text{ pulls})\approx0.2099$

$P(\text{at least }3\text{ reds in }5\text{ pulls})\approx21\%$

In the same way, the probability of pulling at most $3$ red marbles would be the probability of pulling $1$, or $2$, or $3$ red marbles, all added together.

If we’re calculating the probability of at least one success or at least one failure, we can use these formulas:

$P(\text{at least 1 success})=1-P(\text{all failures})$

$P(\text{at least 1 failure})=1-P(\text{all successes})$

This is because all probability distribution functions must add up to $1$.

How to calculate the probability of “at least” and “at most” events

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Probability of getting at least 1 “heads” on 5 coin flips

Example

Find the probability that we get at least $1$ heads on $5$ coin flips.

We can actually simplify this problem a lot by realizing that every single set of $5$ coin flips will have at least one heads, unless every one of the $5$ flips is tails: $TTTTT$. The probability of getting $5$ tails in a row is

$P(TTTTT)=\left(\frac12\right)\left(\frac12\right)\left(\frac12\right)\left(\frac12\right)\left(\frac12\right)=\frac{1}{32}$

The probability of getting “at least one heads” is the same as the probability of not getting “all tails.” Therefore, since total probability is always equal to $1$, we can say that the probability of at least one heads is

$P(\text{at least 1 heads})=1-\frac{1}{32}=\frac{31}{32}$

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Let’s do another example where we find an “at most” probability for a binomial random variable.

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The probability of getting “at least one heads” is the same as the probability of not getting “all tails.”

Example

Let $X$ be a binomial random variable with $n=10$ and $p=0.30$. Find $P(X\le 5)$.

The variable $X$ follows a binomial distribution, but instead of finding the probability of exactly $k$ successes in $n$ trials, we’re asked to find the probability of $k$ or fewer successes in $n$ trials. Specifically, find the chance of $5$ or fewer successes in $10$ trials, where the probability of success on any one trial is $p=0.30$.

Find the probability of $0$ successes, $1$ success, $2$ successes, etc., up to $5$ successes, and then find the sum of those probabilities.

$P(X\le 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$

To find the probability for each value of $k$, we use the binomial probability formula.

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

So the probability $P(X\le 5)$ is

$P(X\le 5)=\binom{10}{0}(0.30)^0(1-0.30)^{10}+\binom{10}{1}(0.30)^1(1-0.30)^{9}$

$+\binom{10}{2}(0.30)^{2}(1-0.30)^8+\binom{10}{3}(0.30)^{3}(1-0.30)^7$

$+\binom{10}{4}(0.30)^{4}(1-0.30)^6+\binom{10}{5}(0.30)^{5}(1-0.30)^5$

$P(X\le 5)=0.9527$

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Mean, variance, and standard deviation

The mean of a binomial random variable $X$ can be expressed as $\mu_X$. The mean is also called the expected value, and that’s indicated as $E(X)$. Either way, the mean is given by

$\mu_X=E(X)=np$

where $n$ is the fixed number of independent trials, and $p$ is the probability of a success. The variance of a binomial random variable $X$ is given by

$\sigma_X^2=np(1-p)$

Standard deviation is the square root of the variance and is therefore given by

$\sqrt{\sigma_X^2}=\sqrt{np(1-p)}$

$\sigma_X=\sqrt{np(1-p)}$

If we continue with our example of the number of heads we get on $5$ coin flips, we can say that the number of trials $n$ is $5$, and the probability of success (getting heads) is $p=0.5$. Therefore, the mean is

$\mu_X=np$

$\mu_X=5(0.5)$

$\mu_X=2.5$

The variance is

$\sigma_X^2=np(1-p)$

$\sigma_X^2=5(0.5)(1-0.5)$

$\sigma_X^2=2.5(1-0.5)$

$\sigma_X^2=2.5(0.5)$

$\sigma_X^2=1.25$

And the standard deviation is

$\sigma_X=\sqrt{np(1-p)}$

$\sigma_X=\sqrt{1.25}$

$\sigma_X\approx1.12$

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