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# Binomial Distribution Word Problems

When an experiment is such that each trial has exactly two possible outcomes, then the probability distribution of such an experiment would be called a binomial probability distribution. A simple example of such an experiment would be to toss a coin $10$ times. For each trial, there are two possible outcomes: heads or tails. Thus, the probability distribution of this experiment would be a binomial probability distribution. In case of a fair coin, the probability of both heads and tails would be equal. However, it is not always so in case of binomial distribution. The probabilities of each outcomes may be different. However, they should add up to $1$. Normally we classify the outcomes as ‘success’ and ‘failures’ depending on what we define as success. The converse would be the failure.

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## Formula

The binomial probability distribution formula is like this:

$P(x)$ = $\frac{N!}{x!(N - x)!}$ $p^x(1 - p)^{N - x}$

Here,

$N$ = Total number of trials in the experiment

$x$ = Total number of successes that we are interested in

$p$ = Probability of success in a single trial

$P$ = Probability of $x$ successes out of $N$ trials

The mean of a binomial distribution is given by the formula:

$\mu$ = $Np$

The variance of a binomial distribution is given by the formula:

$\sigma^2$ = $Np(1 - p)$

So obviously the standard deviation of a binomial distribution would be:

$\sigma$ = $\sqrt{Np(1 - p)}$

## Word Problems

Example 1:

Find the probability that the number of heads exceeds the number of tails when a coin is tossed $6$ times.

Solution:

For this problem,

Number of trials = $N$ = $6$

Probability of success (heads)in each trial = $\frac{1}{2}$ = $0.5$

We are interested in $4$ or more heads, since the number of heads has to be more than tails. So our $x$ would equal $4, 5$ and $6$. Thus using the binomial probability formula we have:

$P(x = 4)$ = $\frac{N!}{x!(N - x)!}$ $p^x$ $(1 - p)^{N - x}$

$P(x = 4)$ = $\frac{6!}{4!(6 - 4)!}$ $0.5^4\ (1 - 0.5)^{6 - 4}$

$P(x = 4)$ = $15 \times 0.5^4 \times 0.5^2$

$P(x = 4)$ = $0.234$

Similarly now let us calculate for $x$ = $5$. Thus we have,

$P(x = 5)$ = $\frac{6!}{5!(6 - 5)!}$ $(0.5)^5$ $(1 - 0.5)^{6 - 5}$

$P(x = 5)$ = $6 \times 0.5^5 \times 0.5^1$

$P(x = 5)$ = $0.094$

Lastly for $x$ = $6$ the probability would be:

$P(x = 6)$ = $\frac{6!}{6!(6 - 6)!}$ $(0.5)^6$ $(1 - 0.5)^{6 - 6}$

$P(x = 6)$ = $1 \times (0.5)^6\ (0.5)^0$

$P(x = 6)$ = $0.016$

Thus the total probability of more heads than tails would be:

= $P(x = 4)+P(x = 5)+P(x = 6)$

= $0.234 + 0.094 + 0.016$

= $0.344 \leftarrow\ Answer!$
Example 2:

Max Life insurance sells $5$ insurance policies to $5$ people all aged $40$ years. According to their years of study data, they know that the probability of a person living to be $70$ or more is $\frac{2}{3}$. Calculate the following probabilities:

a) Exactly $3$ of them live to be $70$ or more.

b) At least $3$ of them are alive after $30$ years.

c) At most $3$ are alive after $30$ years.

Solution:

For this problem, unlike the previous one, the probability of success is $\frac{2}{3}$ and not $\frac{1}{2}$. Thus,

$p$ = $\frac{2}{3}$

The total number of people involved is $5$, so,

$N$ = $5$

a) For this part, we need exactly $3$ to be alive. So our $x$ = $3$. Thus the probability would be:

$P(x = 3)$ = $\frac{N!}{x!(N - x)!}$ $p^x$ $(1 - p)^{N - x}$

$P(x = 3)$ = $\frac{5!}{3!(5 - 3)!}$ $(\frac{2}{3})^3$ $(1 - \frac{2}{3})^{5 - 3}$

$P(x = 3)$ = $10$ $(\frac{8}{27})$ $(\frac{1}{3})^2$

$P(x = 3)$ = $0.329\ \leftarrow\ Answer!$

b) In this case it says at least three are alive. That means either $3$ or $4$ or all $5$ are alive. So here we’d find the probabilities for each of those $x$ values and add them up.

$P(x = 3)$ = $0.329$ (found in part(a))

$P(x = 4)$ = $\frac{5!}{4!(5-4)!}$ $(\frac{2}{3})^4$ $(1 - \frac{2}{3})^{5 - 4}$

$P(x = 4)$ = $5$ $(\frac{16}{81})$ $(\frac{1}{3})^1$

$P(x = 4)$ = $0.329$

$P(x = 5)$ = $\frac{5!}{5!(5 - 5)!}$ $(\frac{2}{3})^5$ $(1 - \frac{2}{3})^{5 - 5}$

$P(x = 5)$ = $1$ $(\frac{32}{243})$ $(\frac{1}{3})^0$

$P(x = 5)$ = $0.132$

Thus the total probability that at least three are alive would be:

$P(x \geq 3)$ = $P(x = 3)+P(x = 4)+P(x = 5)$

$P(x \geq 3)$ = $0.329 + 0.329 + .0132$

$P(x \geq 3)$ = $0.79 \leftarrow\ Answer!$

c) This time it says at most three are alive after $70$ years. So that means, either $0$ or $1$ or $2$ or at most $3$ are alive. Thus let us first find the individual probabilities of each of these.

$P(x = 0)$ = $\frac{5!}{0!(5 - 0)!}$ $(\frac{2}{3})^0$ $(1 - \frac{2}{3})^{5 - 0}$

$P(x = 0)$ = $1$ $(\frac{2}{3})^0$ $(\frac{1}{3})^5$

$P(x = 0)$ = $0.004$

Type equation here.

$P(x = 1)$ = $\frac{5!}{1!(5 - 1)!}$ $(\frac{2}{3})^1$ $(1 - \frac{2}{3})^{5 - 1}$

$P(x = 1)$ = $5$ $(\frac{2}{3})$ $(\frac{1}{3})^4$

$P(x = 1)$ = $0.041$

Type equation here.

$P(x = 2)$ = $\frac{5!}{2!(5 - 2)!}$ $(\frac{2}{3})^2$ $(1 - \frac{2}{3})^{5 - 2}$

$P(x = 2)$ = $10$ $(\frac{2}{3})^2$ $(\frac{1}{3})^3$

$P(x = 2)$ = $0.165$

Type equation here.

$P(x = 3)$ = $0.329$

Thus the combined probability would be:

$P(x \leq 3)$ = $P(x = 0)+P(x = 1)+P(x = 2)+P(x = 3)$

$P(x \leq 3)$ = $0.004 + 0.041 + 0.165 + 0.329$

$P(x \leq 3)$ = $0.539 \leftarrow\ Answer!$
Example 3:

At an archery athletics event in commonwealth games, the probability that George hits the bull’s eye is $(\frac{1}{5})$. If he attempts the shot $8$ times, then find the probability that:

a) He hits the bull’s eye exactly $5$ times.

b) He hits the bull’s eye at least once.

Solution:

In this case the probability of success given to us in the question is:

$p$ = $\frac{1}{5}$

The number of shots that he attempts is $8$, thus,

$n$ = $8$

a) Here we want to find the probability that he hits the bull’s eye exactly $5$ times. Thus here our $x$ = $5$. So the probability using the binomial formula would be:

$P(x)$ = $\frac{N!}{x!(N - x)!}$ $p^x$ $(1-p)^{N-x}$

$P(x = 5)$ = $\frac{8!}{5!(8-5)!}$ $(\frac{1}{5})^5$ $(1 - \frac{1}{5})^{8 - 5}$

$P(x = 5)$ = $56$ $(\frac{1}{5})^5$ $(\frac{4}{5})^3$

$P(x = 5)$ = $0.009\ \leftarrow\ Answer!$

b) In this case we need to find the probability that he hits the bull’s eye at least once. That means we need to find the probability that he hits the bull’s eye once, twice, thrice, $4$ times, $5$ times, $6$ times, $7$ times or $8$ times and then add up all those probabilities. Now that is a very long method! The shorter method would be to find the probability that he does not hit the bull’s eye at all and then subtract that probability from $1$. This would work because the events $x$ = $0$ and $x > 0$ are complimentary events. The total probability of both these events has to be $1$. Also, $P(x > 0)$ is same as $P(x \geq 1)$ since $x$ is a discrete variable here. So let us first find $P(x = 0)$.

$P(x = 0)$ = $\frac{N!}{x!(N - x)!}$ $p^x$ $(1-p)^{N - x}$

$P(x = 0)$ = $\frac{8!}{0!(8 - 0)!}$ $(\frac{1}{5})^0$ $(1 - \frac{1}{5})^{8 - 0}$

$P(x = 0)$ = $1(1)$ $(\frac{4}{5})^8$

$P(x = 0)$ = $0.168$

Therefore,

$P(x \geq 1)$ = $P(x > 0)$ = $1 - 0.168$ = $0.832\ \leftarrow\ Answer!$
Example 4:

Texas state lottery has a prize money of $\$1000.00$. The probability of winning this prize money is$0.35$. If$12$tickets to the lottery are bought, then find the expected winning amount and the standard deviation of the distribution. Solution: The expected value of a binomial experiment is same as its mean. The formula for the mean of a binomial experiment is:$E(x)$=$\mu$=$np$For our problem, there are$12$trials. Thus our$n$=$12$. The probability of success given to us is$0.35$. So the$p$=$0.35$. Thus our mean would be:$\mu$=$12 \times 0.35$=$4.2$Thus the expected winning amount would be:$4.2 \times 1000$=$ \$4200.00\ \leftarrow\ Answer!$

The standard deviation of a binomial distribution is given by the formula:

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

So for this problem it would be:

$\sigma$ = $\sqrt{12 (0.35)(1 - 0.35)}$

$\sigma$ = $1.652$

Thus the standard deviation in terms of number of dollars would be:

$1.652 \times 1000$ = $\$1652.00\ \leftarrow\ Answer!\$
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