Math for Liberal Arts

1. The size of the sample space is \(n\left(S\right) = 6 \cdot 6 = 36\text{.}\) There is only one way we can get a total of 2—if both dice show a 1. The size of the event space is \(n\left(E\right) = 1\text{.}\)

   \begin{equation*} P\left(E\right) = \frac{1}{36} \end{equation*}

2. The size of the sample space is \(n\left(S\right) = 6 \cdot 6 = 36\text{.}\) There are five ways we can get a total of 6:

   \begin{equation*} \{1, 5\}, \{2, 4\}, \{3, 3\}, \{4, 2\}, \{5, 1\} \end{equation*}
   \begin{equation*} n\left(E\right) = 5 \end{equation*}
   \begin{equation*} P\left(E\right) = \frac{5}{36} \end{equation*}

3. It is impossible to get a total of 15—the largest possible total would be \(6 + 6 = 12\text{.}\)

   \begin{equation*} P\left(E\right) = 0 \end{equation*}

1. \begin{equation*} P\left(E^c\right) = 1 - 0.4 = 0.6 \end{equation*}

2. Someone must win the race, so the probability must add up to 1.

   \begin{equation*} P\left(A\right) + P\left(B\right) + P\left(C]\right) + P\left(D\right) = 1 \end{equation*}
   We know that P(A) = 0.4 and the other probabilities are the same. Let P(B) = P(C) = P(D) = x,
   \begin{equation*} 0.4 + x + x + x = 1 \end{equation*}
   \begin{equation*} 0.4 + 3x = 1 \end{equation*}
   \begin{equation*} 3x = 0.6 \end{equation*}
   \begin{equation*} x = 0.6 \div 3 = 0.2 \end{equation*}
   \begin{equation*} P\left(B\right) = 0.2 \end{equation*}

Each roll is independent. The probability of the first roll showing an even number is:

The probability of the 2nd roll showing an even number is also \(\frac{1}{2}\text{.}\)

So the probability of both die showing even numbers is:

The probability of the first pitch being a ball is, P(ball) = 0.3.

Each pitch is an independent event, so the probability of all five being balls is:

1. There are 13 hearts in a deck of cards,

   \begin{equation*} P\left(\text{heart}\right) = \frac{13}{52} = \frac{1}{4} \end{equation*}

2. There is only one King of hearts in the deck,

   \begin{equation*} P\left(\text{King and heart}\right) = \frac{1}{52} \end{equation*}

3. Since we don’t want to count the King of hearts twice, use the following to compute the probability of drawing a King or a heart.

   \begin{equation*} P\left(\text{King or heart}\right) = P\left(\text{King}\right) + P\left(\text{heart}\right) - P\left(\text{King and heart}\right) \end{equation*}
   \begin{equation*} P\left(\text{King}\right) = \frac{4}{52} \end{equation*}
   \begin{equation*} P\left(\text{heart}\right) = \frac{13}{52} \end{equation*}
   \begin{equation*} P\left(\text{King and heart}\right) = \frac{1}{52} \end{equation*}
   \begin{equation*} P\left(\text{King or heart}\right) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \end{equation*}

4. If we know we have drawn a heart, n(S) = 13, because there are 13 hearts in the deck. One of those is a King,

   \begin{equation*} P\left(E\right) = \frac{1}{13} \end{equation*}

1. There are \(_{52}C_{2}\) ways we can draw two cards from a deck.

   \begin{equation*} _{52}C_2 = \frac{52!}{2! 50!} = \frac{52\cdot 51}{2\cdot 1} = 1326 \end{equation*}
   There are \(_{4}C_{2}\) ways we can draw two aces.
   \begin{equation*} _4C_2 = \frac{4!}{2!2!} = \frac{4\cdot 3}{2\cdot 1} = 6 \end{equation*}
   \begin{equation*} P\left(E\right) = \frac{6}{1326} = 0.0045 \end{equation*}

2. There are \(1326\) ways we can draw two cards from a deck (see previous part of the problem). There are \(_{13}C_2\) ways we can draw two hearts.

   \begin{equation*} _{13}C_2 = \frac{13!}{2!11!} = \frac{13\cdot 12}{2\cdot 1} = 78 \end{equation*}
   \begin{equation*} P\left(E\right) = \frac{78}{1326} = 0.0588 \end{equation*}

We can use Bayes theorem or a tree to solve this one.

Suppose we start with a population of 1,000 people. Then \(0.01\left(1000\right) = 10\) people have the disease. The other 990 do not have the disease.

Since the test is 97% accurate, of the 10 people who have the disease:

• 0.97(10) = 9.7 will test positive

• 0.03(10) = 0.3 will test negative

Considering the 990 who do not have the disease:

• 0.97(990) = 960.3 will test negative

• 0.03(990) = 29.7 will test positive

There will be a total of 29.7 + 9.7 = 39.4 positive tests. Only 9.7 of those actually have the disease. Given that a person tests positive, the probability of having the disease is:

The table below gives all possible outcomes, as well as their probabilities.

No, this is not a fair game because the expected value is not 0.

If you put your money in stocks, the following outcomes are possible,

If you put your money in bonds, the following outcomes are possible,

The EV is slightly higher when you invest in stocks, so that would be the better investment.