Suppose you're playing a card game, and you need to draw two aces to win. There are 20 cards left in the deck, and one of the aces has already been drawn. What is the probability that you win the game? How would you go about calculating this probability? Are drawing your first card and drawing your second card independent or dependent events?

### Probability of Compound Events

Recall that **experimental probability** is the ratio of the proposed outcome to the number of experimental trials.

\begin{align*}P(success)= \frac{number \ of \ times \ the \ event \ occured}{total \ number \ of \ trials \ of \ experiment}\end{align*}

Probability can be expressed as a percentage, a fraction, a decimal, or a ratio.

**Compound events** are two simple events taken together, usually expressed as \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.

#### Independent and Dependent Events

Suppose you flip a coin and roll a die at the same time. These are compound events. What is the probability you will flip a head and roll a four?

These events are **independent**. Independent events occur when the outcome of one event does not affect the outcome of the second event. Rolling a four has no effect on tossing a head.

To find the probability of two independent events, multiply the probability of the first event by the probability of the second event.

\begin{align*}P(A \ and \ B)=P(A) \cdot P(B)\end{align*}

\begin{align*}P(tossing \ a \ head)&=\frac{1}{2}\\ P(rolling \ a \ 4)&=\frac{1}{6}\\ P(tossing \ a \ head \ AND \ rolling \ a \ 4)&=\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\end{align*}

When events depend upon each other, they are called **dependent events**. Suppose you randomly draw a card from a standard deck and then randomly draw a second card without replacing the first. The second probability is now different from the first.

To find the probability of two dependent events, multiply the probability of the first event by the probability of the second event, after the first event occurs.

\begin{align*}P(A \ and \ B)=P(A) \cdot P(B \ following \ A)\end{align*}

#### Let's solve the following problem:

Two cards are chosen from a deck of cards. What is the probability that they both will be face cards?

Let \begin{align*}A = 1st \ Face \ card \ chosen\end{align*} and \begin{align*}B = 2nd \ Face \ card \ chosen\end{align*}. The total number of face cards in the deck is \begin{align*}4 \times 3 = 12\end{align*}.

\begin{align*}P(A)&= \frac{12}{52}\\ P(B) & = \frac{11}{51}, \ \text{remember, one card has been removed.}\end{align*}

\begin{align*}P(A \ and \ B)= \frac{12}{52} \times \frac{11}{51} \ & or \ P(A \cap B) = \frac{12}{52} \times \frac{11}{51} =\frac{33}{663}\\ P(A \cap B) & = \frac{11}{221}\end{align*}

Note that \begin{align*}P(A \cap B)\end{align*} is equivalent to \begin{align*}P(A \space and \space B)\end{align*}.

#### Mutually Exclusive Events

Compound events that cannot happen at the same time are called **mutually exclusive** events. For example, a number cannot be both even and odd or you cannot have picked a single card from a deck of cards that is both a ten and a jack. **Mutually inclusive** events, however, can occur at the same time. For example a number can be both less than 5 and even or you can pick a card from a deck of cards that can be a club and a ten.

When finding the probability of events occurring at the same time, there is a concept known as the “double counting” feature. It happens when the intersection is counted twice.

In mutually exclusive events, \begin{align*}P(A \space and \space B)=0\end{align*}, because they cannot happen at the same time.

To find the probability of either mutually exclusive event \begin{align*}A\end{align*} or \begin{align*}B\end{align*} occurring, use the following formula:

\begin{align*}P(A \ or \ B)=P(A)+P(B)\end{align*}

To find the probability of one or the other mutually inclusive event, add the individual probabilities and subtract the probability they occur at the same time.

\begin{align*}P(A \ or \ B)=P(A)+P(B)-P(A \space and \space B)\end{align*}

Note that the formula for finding the probability of one or the other mutually exclusive event is the same as the formula for finding the probability of one or the other mutually inclusive event except \begin{align*}P(A \space and \space B) = 0\end{align*} in the case of the mutually exclusive events.

#### Let's find the following probabilities given the information below:

Two cards are drawn from a deck of cards. Let:

\begin{align*}A\end{align*}: \begin{align*}1^{st}\end{align*} card is a club

\begin{align*}B\end{align*}: \begin{align*}1^{st}\end{align*} card is a 7

\begin{align*}C\end{align*}: \begin{align*}2^{nd}\end{align*} card is a heart

Find the following probabilities:

- \begin{align*}P(A \ \text{or} \ B)\end{align*}

A club or a 7 can be picked at the same time so these are mutually inclusive events. You can use the formula from above.

\begin{align*}& P(A \ or \ B)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\\
& P(A \ or \ B) =\frac{16}{52}\\
& P(A \ or \ B) =\frac{4}{13}\end{align*}

- \begin{align*}P(B \ \text{or} \ A)\end{align*}

A club and a 7 can be picked at the same time so these are mutually inclusive events. You can use the formula from above.

\begin{align*}& P(B \ or \ A)= \frac{4}{52} + \frac{13}{52}-\frac{1}{52}\\
& P(B \ or \ A) = \frac{16}{52}\\
& P(B \ or \ A) = \frac{4}{13}\end{align*}

- \begin{align*}P(A \ \text{and} \ C)\end{align*}

Picking a club on the first card and a heart on the second card are dependent events so you need to multiply the probability of A by the probability of C following A.

\begin{align*}& P(A \ and \ C) = \frac{13}{52} \times \frac{13}{51}\\ & P(A \ and \ C) = \frac{169}{2652}\end{align*}

### Examples

#### Example 1

Earlier, you were told that you are playing a card game and need to draw two aces to win. There are 20 cards left in the deck and one of the aces has already been drawn. What is the probability that you win the game? Are drawing your first card and drawing your second card independent or dependent events?

Since you are not replacing the card that you draw, the event of drawing an ace on your first draw and the event of drawing an ace on your second are dependent events. To find this probability, you need to multiply the probability of getting an ace on the first draw by the probability of getting an ace on the second draw given that you already drew an ace on the first draw.

If there are 20 cards left in the deck and one ace has already been drawn, there are 3 possible aces in a deck of 20.On your first try, there is \begin{align*}\frac{3}{20}\end{align*} chance that you will pick an ace. On the second draw, if you draw an ace in the first draw, there is one less ace and one less card in the deck. The probability of getting another ace on the second draw is \begin{align*}\frac{2}{19}\end{align*}.

\begin{align*}\frac{3}{20}\times \frac{2}{19} = \frac{3}{190}\end{align*}

The probability of getting two aces and winning is \begin{align*}\frac{3}{190}\end{align*}.

#### Example 2

A bowl contains 12 red marbles, 5 blue marbles and 13 yellow marbles. Find the probability of drawing a blue marble and then drawing a yellow marble.

Let \begin{align*}A = blue \ marble \ chosen \ 1st \end{align*} and \begin{align*}B = yellow \ marble \ chosen \ 2nd \end{align*}. The total number of marbles in the bowl is \begin{align*}12+5+13=30\end{align*}.

\begin{align*}P(A)&= \frac{5}{30}\\ P(B) & = \frac{13}{29} \ \text{Remember, one marble has been removed.}\end{align*}

\begin{align*}P(A \ AND \ B)= \frac{5}{30} \times \frac{13}{29} \ & or \ P(A \cap B) =\frac{5}{30} \times \frac{13}{29} =\frac{65}{870}\\ P(A \cap B) & = \frac{13}{174}\end{align*}

### Review

- Define independent events.

Are the following events independent or dependent?

- Rolling a die and spinning a spinner
- Choosing a book from the shelf and then choosing another book without replacing the first
- Tossing a coin six times and then tossing it again
- Choosing a card from a deck, replacing it, and choosing another card
- If a die is tossed twice, what is the probability of rolling a 4 followed by a 5?
- Define mutually exclusive.

Are these events mutually exclusive or mutually inclusive?

- Rolling an even and an odd number on one die.
- Rolling an even number and a multiple of three on one die.
- Randomly drawing one card and getting a result of a jack and a heart.
- Randomly drawing one card and getting a result of a black and a diamond.
- Choosing an orange and a fruit from a basket.
- Choosing a vowel and a consonant from a Scrabble bag.
- Two cards are drawn from a deck of cards. Determine the probability of each of the following events:
- \begin{align*}P\end{align*}(heart or club)
- \begin{align*}P\end{align*}(heart and club)
- \begin{align*}P\end{align*}(red or heart)
- \begin{align*}P\end{align*}(jack or heart)
- \begin{align*}P\end{align*}(red or ten)
- \begin{align*}P\end{align*}(red queen or black jack)

- A box contains 5 purple and 8 yellow marbles. What is the probability of successfully drawing, in order, a purple marble and then a yellow marble? {Hint: In order means they are not replaced.}
- A bag contains 4 yellow, 5 red, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 yellow marbles?
- A card is chosen at random. What is the probability that the card is black and is a 7?

**Mixed Review**

- A circle is inscribed within a square, meaning the circle's diameter is equal to the square’s side length. The length of the square is 16 centimeters. Suppose you randomly threw a dart at the figure. What is the probability the dart will land in the square, but not in the circle?
- Why is \begin{align*}7-14x^4+7xy^5-1x^{-1}=8x^2 y^3\end{align*} not considered a polynomial?
- Factor \begin{align*}72b^5 m^3 w^9-6(bm)^2 w^6\end{align*}.
- Simplify \begin{align*}2^5-7^3 a^3 b^7+3^5 a^3 b^7-2^3\end{align*}.
- Bleach breaks down cotton at a rate of 0.125% with each application. A shirt is 100% cotton.
- Write the equation to represent the percentage of cotton remaining after \begin{align*}w\end{align*} washes.
- What percentage remains after 11 washes?
- After how many washes will 75% be remaining?

- Evaluate \begin{align*}\frac{(100 \div 4 \times 2-49)^2}{9-2 \times 3+2^2}\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 9.12.