Examples of the Binomial Distribution

1 in 5 ride without tickets, which is what you’re being asked to examine. That means that

That makes this a binomial distribution with $n=20$ and $p=0.2$.

1.

First you want to find the probability that exactly one of the 20 doesn’t have a ticket. In that case $k=1$, and the formula becomes $$\begin{array}{llll}\hfill P\left(X=1\right)& =\left(\genfrac{}{}{0.0pt}{}{20}{1}\right)p^1{(1-p)}^{20-1}& \hfill & \\ \hfill & =20\cdot 0.2\cdot 0.8^{19}& \hfill & \\ \hfill & \approx 0.06& \hfill & \end{array}$$

The probability that exactly one of the $20$ doesn’t have a ticket is around $6$ %.

2.

Next, we want to find the probability that exactly five of the 20 don’t have a ticket. In that case $k=5$, which makes the formula $$\begin{array}{llll}\hfill P\left(X=5\right)& =\left(\genfrac{}{}{0.0pt}{}{20}{5}\right)p^5{(1-p)}^{20-5}& \hfill & \\ \hfill & =15504\cdot 0.2^5\cdot 0.8^{15}& \hfill & \\ \hfill & \approx 0.17& \hfill & \end{array}$$

The probability that exactly five of the 20 ride without a ticket is around $17$ %.

3.

Finally, you want to find the probability that at least three of the 20 don’t have a ticket. That means that either 3, 4, 5 and so on up to 20 don’t have a ticket. As this is a lot of outcomes to find the probability of, it can be smart to look at the complement instead. Our inquiry is the same as saying that “0, 1 or 2 out of 20 don’t have a ticket” doesn’t occur after all. This gives you

$$P\left(X\ge 3\right)=1-P\left(X=\text{0, 1 or 2}\right)$$

To calculate $P\left(X=\text{0, 1 or 2}\right)$, you have to add up $P\left(X=0\right),P\left(X=1\right)$ and $P\left(X=1\right)$, making the first order of business to find them using the same formula you used in Items 2 and 3.

The probability of guessing the correct answer to a question with four alternatives is $\frac{1}{4}=0.25=25\text{\%}$.

1.

As the quiz has 50 questions, $n=50$. Stephen needs all of them to be correct to achieve $k=50$. This gives you that

$$P\left(X=k\right)=\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)p^k{(1-p)}^{n-k}$$

becomes

$$\begin{array}{llll}\hfill P\left(X=50\right)& =\left(\genfrac{}{}{0.0pt}{}{50}{50}\right)\cdot 0.25^{50}\cdot {(1-0.25)}^{50-50}& \hfill & \\ \hfill & =1\cdot 0.25^{50}\cdot 0.75^0& \hfill & \\ \hfill & \approx 7.89\cdot 10^{-31}& \hfill & \end{array}$$

Sadly, Stephen is probably never going to Paris.

2.

As the quiz has 50 questions, $n=50$. To achieve 30 correct answers, Stephen needs $k=30$:

$$\begin{array}{llll}\hfill P\left(X=30\right)& =\left(\genfrac{}{}{0.0pt}{}{50}{30}\right)\cdot 0.25^{30}& \hfill & \\ \hfill & \cdot {(1-0.25)}^{20}& \hfill & \\ \hfill & \approx 1.30\cdot 10^{-7}& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill P\left(X=30\right)& =\left(\genfrac{}{}{0.0pt}{}{50}{30}\right)\cdot 0.25^{30}\cdot {(1-0.25)}^{20}& \hfill & \\ \hfill & \approx 1.30\cdot 10^{-7},& \hfill & \end{array}$$

which means that he’s not likely to go to Hawaii either.

3.

To get at least five correct answers means getting between 5 and 50 correct answers. You have to find the probability of $X\ge 5$. This is the same as the probability that $X$ isn’t 0, 1, 2, 3 or 4. That means you can use the rule for complementary events $P\left(A\right)=1-P\left(\stackrel{}{A}\right)$ to calculate

The sum of these probabilities is

If you insert this into the expression above, you get that

It seems like Stephen at least gets to go to the Museum of Natural History for free.