A snack bag of animal crackers contains 5 elephants, 7 monkeys, and 4 lions. What is the probability that the first two crackers you randomly pull from the bag will be a lion followed by an elephant?

### Guidance

When multiple events occur we can calculate the probability of these combined events by finding their product if the events are independent. **Independent Events** are events for which the outcome of one event does not affect the outcome of a second event. For example, if we roll a die and then roll it again, the outcome of the second roll is independent from the outcome of the first event.

To determine the probability of two independent events, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, both occurring, we multiply the probabilities of each of the two events together: \begin{align*}P(A) \times P(B) = P(A \ and \ B)\end{align*}.

\begin{align*}^*\end{align*}Note, the use of set notation will be introduced in the concept *Union and Intersection of Sets.*

In some cases, the outcome of one event affects the outcome of a second event. For example, when a hand of cards is dealt in a game of poker, the probability of receiving a particular card changes based on what cards have already been dealt. This is an example of **Conditional Probability**. We will introduce conditional probability here for situations in which we can manipulate the subsequent probabilities to make independent events as shown in Example C.

#### Example A

Given a fair die and a fair coin, find the following probabilities.

- rolling a 5 and getting tails.
- rolling an odd number and getting heads.

**Solution:** Since the outcome of rolling the die does not affect the outcome of flipping the coin, these are independent events. Thus we can determine their individual probabilities and multiply them together.

- \begin{align*}P(5) \times P(T) = \frac{1}{6}\times\frac{1}{2}=\frac{1}{12}\end{align*}.
- \begin{align*}P(1, 3, 5) \times P(H) = \frac{3}{6}\times\frac{1}{2}=\frac{3}{12}=\frac{1}{4}\end{align*}

#### Example B

What is the probability of rolling doubles twice in a row? Three times in a row?

**Solution:** Each of these rolls are independent events. It is human nature to think that just because we have rolled doubles once or twice already that we are unlikely to roll them another time. It is true that that probability of rolling doubles three times in a row is smaller than rolling doubles once but this is not because the probability changes for each roll. Let’s look at why this occurs:

\begin{align*}P(\text{doubles}) = \frac{6}{36}=\frac{1}{6}\end{align*}, since there are six ways to roll doubles.

\begin{align*}P(\text{doubles twice}) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\end{align*}

\begin{align*}P(\text{doubles three times}) = \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}=\frac{1}{216}\end{align*}

#### Example C

What is the probability that you draw an ace from a deck of cards three times if each card is replaced before the next card is drawn? What if each card is not replaced?

**Solution:** There are 4 aces in a deck of cards, so there is a \begin{align*}\frac{4}{52}=\frac{1}{13}\end{align*} chance of drawing an ace each time a card is selected. For the first part of the question which requires each card chosen to be replaced, the probability of selecting an ace does not chance so the events are independent of one another.

\begin{align*}P(\text{three aces, with replacement}) = \left(\frac{1}{13}\right)^3=\frac{1}{2197}\end{align*}.

The second part of the question does not require replacement. Now the events are not strictly independent. We can, however, determine the probability of an independent event and use multiplication to find the probability of the combined events. For the first selection, there are 4 aces in the 52 card deck. After an ace is selected, how many cards remain? Well, to determine the probability of selecting three aces, we must assume the first selected was an ace so now there are 3 aces remaining in a deck of 51 cards. After the second ace is selected, there are 2 aces in a deck of 50 cards. Now we can find the product of these probabilities.

\begin{align*}P(\text{three aces, without replacement}) = \frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}=\frac{1}{13}\times\frac{1}{17}\times\frac{1}{25}=\frac{1}{5525}\end{align*}.

Notice that the probability of selecting an ace diminishes with each selection in this situation because the number of aces in the deck is being reduced.

**Intro Problem Revisit** There are \begin{align*}7+5+4=16\end{align*} crackers in the bag. The probability that the first cracker you pull will be a lion will therefore be \begin{align*}\frac{4}{16} = \frac{1}{4}\end{align*}.

Now there are 15 crackers remaining in the bag, so the probability that the second cracker you pull from the bag will be an elephant is \begin{align*}\frac{5}{15}=\frac{1}{3}\end{align*}.

Finally, \begin{align*}P(\text{lion followed by elephant, without replacement}) = \frac{1}{4}\times\frac{1}{3}=\frac{1}{12}\end{align*}.

### Guided Practice

Determine the following probabilities.

1. Rolling snake eyes (double ones) and then rolling a sum of seven on a pair of dice.

2. Turning over a face card (jack, queen or king) followed by an ace from a full, shuffled deck of cards.

3. Drawing a hand of five cards which contains exactly 2 jacks, from a full, shuffled deck of cards.

4. Randomly selecting a pair of black socks followed by a pair of navy blue socks and then a pair of white socks from a drawer containing 5 black pairs, 4 navy blue pairs and 8 white pairs if each selection is replaced before the next pair is chosen. What if each pair is not replaced?

#### Answers

1. \begin{align*}P(\text{snake eyes}) \times P(\text{seven}) = \frac{1}{36}\times\frac{1}{6}=\frac{1}{216}\end{align*}.

2. \begin{align*}P(\text{jack, queen, king}) \times P(\text{ace}) = \frac{12}{52}\times\frac{4}{51}=\frac{4}{221}\end{align*}

3. \begin{align*}P(\text{jack}) \times P(\text{jack}) \times P(\text{non jack}) \times P(\text{non jack}) \times P(\text{non jack}) =\frac{4}{52}\times \frac{3}{51} \times \frac{48}{50} \times \frac{47}{49}\times \frac{46}{48}=\frac{1081}{270725}\approx 0.00399298\end{align*}

We aren’t quite finished, however, because what we have determined here is the probability that the first two cards are the jacks and the last three are the non jacks. These cards could have been dealt in any order so we need to determine the number of permutations of these cards and multiply by that value. Keep in mind that the jacks are “non distinguishable” and the non jacks are “non distinguishable”. The permutations are: \begin{align*}\frac{5!}{2!3!}=10\end{align*}. This could also be described as the number of combinations of selecting two jacks from a set of five cards: \begin{align*}_5C_2=\frac{5!}{2!(5-2)!}=10\end{align*}. These are just two ways to find the same result. Now that we have the number of combinations or arrangements, we can multiple our probability by this value:

\begin{align*}\frac{1081}{270725}\times10\approx0.0399298181\end{align*}.

4. For the first part of the question, the total number of socks in the drawer remains the same for each selection, 17. So, \begin{align*}P(\text{black, then blue, then white}) = \frac{5}{17}\times\frac{4}{17}\times\frac{8}{17}=\frac{160}{4913}\approx0.032567\end{align*}. Now, if we do not replace the socks, the number of socks decreases with each sock selected: \begin{align*}P(\text{black, then blue, then white}) = \frac{5}{17}\times\frac{4}{16}\times\frac{8}{15}=\frac{160}{4080}=\frac{2}{51}\approx0.039216\end{align*}. This probability is slightly higher because removing one pair of black socks makes it more likely that we will select a different color pair of socks for the next pair and so on.

### Explore More

Given a spinner as shown in the illustration below, two fair six sided dice and a standard deck of 52 playing cards, calculate the probability of each compound event below.

- Spin a 4 and roll doubles.
- Spin an odd number and roll an odd sum.
- Draw four red cards in a row without replacement.
- Draw three face cards (jack, queen, king) without replacement.
- Rolling a sum of eight and spinning a 2.
- Spin a three, three times in a row.
- Roll an even sum and spin a prime number.
- Draw a five card hand containing exactly 3 red and 2 black cards.
- Draw a five card flush (a hand in which all cards are the same suit).
- Draw a hand of five cards containing exactly two hearts.
- Draw a hand of three cards that contains at least one spade.
- Roll a sum of 7 or 11 and draw three cards in which at least one is a face card.