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# Independent/Dependent Events

In probability, the sets of outcomes of an experiment are called events. There are a variety of types of events, including dependent events, independent events, mutually exclusive events, and so on. For now, we are focusing on dependent and independent events.

## What are independent events?

Two or more events are considered independent if the results of the events following the first event are not affected by the result of any of the preceding events. As an example, if you flip a coin 20 times, you will come up with a certain number of heads and a certain number of tails. However, no individual flip has anything to do with the result of another flip. Each flip is an independent event.

We could say that if the probability of event A is not affected by the occurrence of another event B, then A and B are independent events.

## Determining the probability of independent events

To find the probability, or odds, of independent events, you simply multiply the probability of each of the individual events. This can be written as:

$P\left(AandB\right)=P\left(A\right)×P\left(B\right)$

Let's look at this in action.

Example 1

Let's say you have a bag with 5 green marbles, 3 blue marbles, and 2 red marbles. For each event, one marble is removed from the bag, the color is recorded, and then the marble is replaced. Then another marble is chosen. What is the probability that the first marble is red and the second marble is green?

Because the first marble is replaced, the number of marbles we can draw (10) does not change from the first drawing to the second drawing. That makes each drawing an independent event.

$P\left(\mathrm{red}then\mathrm{green}\right)=P\left(\mathrm{red}\right)×P\left(\mathrm{green}\right)$

$=\frac{2}{10}×\frac{5}{10}$

$=\frac{10}{100}$

$=\frac{1}{10}$

So the odds are one in ten that the first draw will be a red and the second draw will be a green.

What happens when we do four draws instead of two? What is the probability that we get a blue marble on the first draw, a green marble on the second draw, a red marble on the third draw, and a blue marble on the fourth draw?

Example 2

$P\left(\mathrm{blue},\mathrm{green},\mathrm{red},\mathrm{blue}\right)=P\left(\mathrm{blue}\right)×P\left(\mathrm{green}\right)×P\left(\mathrm{red}\right)×P\left(\mathrm{blue}\right)$

$=\frac{3}{10}×\frac{5}{10}×\frac{2}{10}×\frac{3}{10}$

$=\frac{90}{10000}$

$=\frac{9}{1000}$

So your odds are 9 in 1000 that you will pull out four marbles in the exact order given.

## What are dependent events?

Events are dependent when the occurrence of one event affects the probability of the occurrence of the other. Dependent events are affected by the outcomes that have already occurred previously. If one event is changed, then another is likely to differ.

For example, if you are drawing cards from a deck and not returning the card, the probability of a particular card changes each time you draw a new card because the sample pool has been lessened by one card.

## Determining the probability of dependent events

When two events, A and B, are dependent, the probability of the outcome of A and B is:

$P\left(A\mathrm{and}B\right)=P\left(A\right)×P\left(A|B\right)$

Let's see how this works with the same bag of marbles.

Example 3

Again, you have a bag with 5 green, 3 blue, and 2 red marbles. This time, when you draw a marble from the bag, you do not replace it. What is the probability that you draw first a red and then a green marble?

$P\left(\mathrm{red}\mathrm{then}\mathrm{green}\right)=P\left(\mathrm{red}\right)×P\left(\mathrm{green}|\mathrm{red}\mathrm{first}\right)$

The probability of the first event happening is the same as if it were an independent event. The second probability is now 5 out of 9, since the bag has only 9 marbles left in it.

$\frac{2}{10}×\frac{5}{9}=\frac{10}{90}=\frac{1}{9}$

So the odds of getting one red marble and then one green marble actually improve to 1 in 9 when they become independent events. This is because there are more green marbles than red marbles, so you are more likely to draw a green marble the second time when one red marble has been removed.

What about the likelihood of getting one blue, then one green, then one red, and then another blue marble?

Example 4

$P\left(\mathrm{blue},\mathrm{green},\mathrm{red},\mathrm{blue}\right)=P\left(\mathrm{blue}\right)×P\left(\mathrm{green}|\mathrm{blue}\right)×P\left(\mathrm{red}|\mathrm{green}\mathrm{and}\mathrm{blue}\right)×P\left(\mathrm{blue}|\mathrm{green},\mathrm{blue},\mathrm{and}\mathrm{red}\right)$

Let's see how this will work out.

The chances of drawing a blue marble first are $\frac{3}{10}$ because you have a full bag of marbles.

The chances of drawing a green marble second are $\frac{5}{9}$ , because you still have all the green marbles and only 9 marbles are left in the bag.

The chances of drawing a red marble third are $\frac{2}{8}$ , because you still have both red marbles and only 8 marbles are left in the bag.

The chances of drawing a blue marble fourth are $\frac{2}{7}$ , because you only have 2 blue marbles left and there are only 7 marbles in the bag.

So the probability of this draw is:

$\frac{3}{10}×\frac{5}{9}×\frac{2}{8}×\frac{2}{7}$

$\frac{1}{84}$

So the chances of drawing the marbles in the given order when they are not replaced back in the bag are 1 in 84.

Let's look at another example of dependent events.

Example 5

In a pack of 52 cards, a card is drawn at random without being replaced. Find the probability of drawing a jack of spades followed by drawing a red card.

Because the first card is not replaced, these two events are dependent. So we will use:

$P\left(A\mathrm{and}B\right)=P\left(A\right)×P\left(A|B\right)$

$P\left(\mathrm{jack of spades}\mathrm{then}\mathrm{a red card}\right)=P\left(\mathrm{jack of spades}\right)×P\left(\mathrm{red card}|\mathrm{jack of spades}\right)$

The probability of drawing a jack of spades is $\frac{1}{52}$ .

There are 26 red cards, so the probability of drawing a red card is now $\frac{26}{51}$ .

The chances of drawing these two cards in order are:

$\frac{1}{52}×\frac{26}{51}$

$\frac{26}{2652}$

or $\frac{1}{102}$ .

## Flashcards covering the Independent/Dependent Events

Statistics Flashcards