## Independent Events

Two outcomes are ** independent**Â if the occurrence of one outcome has no effect on the occurrence of the other. For example, if a coin is tossed twice, the first outcome (H or T) has no effect on the second outcome (H or T).

There are two ways multiple events can take place in a single probability problem: either they can each occur separately (AÂ *or*Â B) or they must occur together (AÂ *and*Â B).

## “Or”

In some scenarios, the events do not have to occur together to have the desired result. One of the events is enough.

*I will be happy today if I win the lottery OR have email.*

“Or” means that either outcome is desired. With more possible desired outcomes, the probability is greater than for one event alone.

For independent events with “or,”Â addÂ the probabilities of the events.

### Solution

First, notice the word “or” in the question. There are 6 possible outcomes, and either a 5Â ** or**Â a 6 is wanted. There is a 1 in 6 probability of a 5 and a 1 in 6 probability of a 6.

**Bottom:**Â 6 outcomes are possible

**Top:**Â 2 possible desired outcomes

**Probability:** \dfrac{2}{6} = \dfrac{1}{3}

### Solution

John can win by pulling out either a 7Â ** or**Â a 9. His chance of doing this is higher than if he could only win by pulling out a 7. In that case, he’d only have 4 cards that would win, in this case there are 8 winning cards.

*Method 1*

To find the total probability, figure out the probability of each event and then add probabilities.

Pulling a 7

**Bottom:**Â 52 outcomes are possible

**Top:**Â 4 possible desired outcomes

**Probability:** \dfrac{4}{52} = \dfrac{1}{13}

Pulling a 9

**Bottom:**Â 52 outcomes are possible

**Top:**Â 4 possible desired outcomes

**Probability:**Â \dfrac{4}{52} = \dfrac{1}{13}

Probability of pulling a 7 or a 9:

\dfrac{1}{13} + \dfrac{1}{13} = \dfrac{2}{13}

*Method 2*

Because the total possible outcomes are the same for both events, there is no need to calculate their probabilities separately.

**Bottom:**Â 52 outcomes are possible

**Top:**Â 8 possible desired outcomes

**Probability:** \dfrac{8}{52} = \dfrac{2}{13}

### Solution

This is an “or” question, even though the word “or” isn’t written. The probability is independent. Think about what it means to have heads come up only once. Be careful to recognize that the desired outcome is heads only once, not at least once. Write the possible desired scenarios.

**H T T**Â orÂ **T H T**Â orÂ **T T H**

There are three outcomes that would achieve the desired result.

**Probability of H T T** = \dfrac{1}{2} Ã— \dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{8}

**Probability of T H T** = \dfrac{1}{2} Ã— \dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{8}

**Probability of T T H** = \dfrac{1}{2} Ã— \dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{8}

Since these are independent events, add.

\dfrac{1}{8} Ã— \dfrac{1}{8} Ã— \dfrac{1}{8} = \dfrac{3}{8}### Solution

The probabilities are independent. Each event has a different number of possible outcomes. The only way to solve this type of problem, where the events have a different number of possible outcomes, is to add the separate probabilities.

Getting a heads

**Bottom:**Â 2 outcomes are possible

**Top:**Â 1 possible desired outcome

**Probability:** \dfrac{1}{2}

Rolling a 3

**Bottom:**Â 6 outcomes are possible

**Top:**Â 1 possible desired outcome

**Probability:** \dfrac{1}{6}

Probability of getting a heads or rolling a 3: \dfrac{1}{2} + \dfrac{1}{6} = \dfrac{3}{6} + \dfrac{1}{6}Â

= \dfrac{4}{6} = \dfrac{2}{3}

## “And”

In some scenarios, all the events have to occur in order to attain the desired result.

*I will be happy today if I win the lottery AND have email.*

“And” means that both outcomes are needed. With more outcomes required, the probability is less.

For independent events with “and,”Â multiplyÂ the probabilities of the events.

#### 800score Tip

In general, the probability of many events occurring is less than just one event occurring.

Probability is expressed as a fraction between 0 and 1. When you multiply these fractions, you get *smaller*Â fractions, as in “AND” scenarios. When you add these fractions you get *greater* fractions, as in “OR” scenarios.

### Solution

First, notice the word “and” in the question. That means both events must occur. So the desired coin toss isÂ **H T**.

The probability that the coin will land on heads is \dfrac{1}{2}. The probability that the coin will land on tails is also \dfrac{1}{2}.

Since both must happen, multiply the individual probabilities.

\dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{4}As expected, the probability of both events occurring is less than the probabilities of either individual event occurring.

Not all “and” questions will include the word “and.” It may say “then” or “both” or it may just be implied.

### Solution

In this case, the first die must be a 5 and the second die must also be a 5. This is an implied “and.”

Probability of a 5 on the first die is \dfrac{1}{6}

Probability of a 5 on the second die is \dfrac{1}{6}

Probability both are 5 is \dfrac{1}{6} Ã— \dfrac{1}{6}Â

= \dfrac{1}{36}

## With Replacement

Sometimes independent event questions use the term “with replacement.” This means that the first card or marble that is chosen is put back or replaced, so it could be chosen again and not affect the probability of subsequent events (i.e., all the events are independent).

### Solution

Since both tokens must be red, this is an “and” question. Since the events are the same and the first token is replaced, they will each have the same probability. Calculate the probability then multiply.

**Bottom:**Â 52 outcomes are possible

**Top:**Â 13 tokens are red

**Probability:** \dfrac{13}{52} = \dfrac{1}{4}

Probability of pulling 2 reds:

\dfrac{1}{4} Ã— \dfrac{1}{4} = \dfrac{1}{16}

### Solution

The “then” in the question implies “and.” Since both outcomes must occur, and there is replacement, these are independent events. Multiply the probabilities.

Probability of a red marble

**Bottom:**Â 10 outcomes are possible

**Top:**Â 5 marbles are red

**Probability:** \dfrac{5}{10} = \dfrac{1}{2}

Probability of a blue marble

**Bottom:**Â 10 outcomes are possible

**Top:**Â 2 marbles are blue

**Probability:** \dfrac{2}{10} = \dfrac{1}{5}

Probability of red then blue:

\dfrac{1}{2} Ã— \dfrac{1}{5} = \dfrac{1}{10}