Probability of Independent Events

 

Learning Objective(s)

·         Calculate the probability of independent events.

 

 

Some probability situations involve more than one event. When the events do not affect one another, they are known as independent events. Independent events can include repeating an action like rolling a die more than once, or using two different random elements, such as flipping a coin and spinning a spinner. Many other situations can involve independent events as well. In order to calculate probabilities accurately, we need to know if one event influences the outcome of other events.

 

 

The main characteristic of a situation with independent events is that the original state of the situation doesn't change when one event occurs. There are two ways in which this happens:

 

Independent events occur when either   ·         the process that generates the random element doesn't remove any possible outcomes, or ·         the process does remove a possible outcome, but the outcome is replaced before a second action takes place. (This is referred to as drawing with replacement.)

 

 

Here are examples of each case:

 

 

 

Let's look more closely at the second example. On the first pull, the probability of getting red is , because there are 5 marbles and 2 of them are red. If that red marble is replaced (put back in the bag),  the probability for the second pull is still , and that means the events are independent. The result of one trial does not affect the result of another.

 

But what would happen if the first marble wasn't put back in the bag? Then the probability of getting red will be different for the second pull. If one red marble is removed, on the second pull the probability is now  because there are only 4 marbles and 1 is red.

 

Now let's go back to the first example. Suppose a die were rolled 15 times without getting a 6. On the next roll, is the probability of rolling a 6 still , or is it greater? Some people may believe the next roll is more likely to be a 6, because "I'm due for a 6!" However, this is incorrect—the die cannot remember what was rolled before. While it's somewhat unusual to roll 16 times without getting a 6, the probability of getting a 6 after 15 of those rolls have gone by is the same as getting a 6 on any other roll.

 

Latonya is playing a card game. She starts with 10 cards, numbered from 1 to 10, face down so she can’t see the numbers. She randomly chooses a card and turns it over. If the card is greater than 5, it's a "win" and she puts it in the "win" pile. If the card is 5 or less, she puts it in the "lose" pile. She wins the game if she gets three cards in the win pile before she gets three in the lose pile. She wants to calculate the probability of winning the game with the first three cards.   Choose the sentence that best matches this situation.   A) The events are independent, because the game doesn't remove any outcomes. B) The events are independent, because each round has the same two possible outcomes (win or lose). C) The events are not independent, because an outcome is removed at each turn and not replaced.   Show/Hide Answer A) Incorrect. With each turn, Latonya removes a card from the 10 she starts with. One of the possible outcomes is removed and the probability of getting a "win" changes each turn. The events are not independent.   B) Incorrect. Because Latonya's card goes into the "win" or "lose" pile, it is not replaced, so the probability of getting a "win" changes each turn. The events are not independent.   C) Correct. Latonya removes a card each turn and does not replace it, so the probability of getting a "win" changes each turn.

 

 

 

Let's look at the sample and event spaces for the examples given in the previous section.

 

·         You roll a die twice. What's the probability of getting a 6 on the second roll but not on the first?

 

In this example, a die is rolled twice.

		First roll
		1	2	3	4	5	6
Second roll	1	1,1	2,1	3,1	4,1	5,1	6,1
	2	1,2	2,2	3,2	4,2	5,2	6,2
	3	1,3	2,3	3,3	4,3	5,3	6,3
	4	1,4	2,4	3,4	4,4	5,4	6,4
	5	1,5	2,5	3,5	4,5	5,5	6,5
	6	1,6	2,6	3,6	4,6	5,6	6,6

 

 

There are 6 possible outcomes for the first roll, and for each of those, there are 6 possible outcomes for the second roll. There are 6 • 6, or 36, possible outcomes:

 

Sample space: {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

 

The event space is all the outcomes for which the first roll was not 6, and the second roll was 6. For the first roll, there were 5 possible outcomes that are not 6. For each of those, there is 1 possible outcome that is a 6. So there are 5 • 1 or 5 outcomes in the event space:

 

Event space: {(1,6), (2,6), (3,6), (4,6), (5,6)}

 

Notice that the size of the sample space for both rolls is the product of the size of the sample space for each roll. Similarly, the size of the event space for two rolls is the product of the size of the event space for each roll.

 

On to scenario 2:

 

·         You pull a marble from a bag with 2 red, 2 white, and 1 green marble. You note the color, put it back, and then pull another marble. What's the probability of pulling red marbles both times?

 

To help remind us that there are two red marbles, let's call them R1 and R2. We'll do the same with the white marbles, W1 and W2.

 

		First pull
		R1	R2	W1	W2	G
Second pull	R1	R1,R1	R2,R1	W1,R1	W2,R1	G,R1
	R2	R1,R2	R2,R2	W1,R2	W2,R2	G,R2
	W1	R1,W1	R2,W1	W1,W1	W2,W1	G,W1
	W2	R1,W2	R2,W2	W1,W2	W2,W2	G,W2
	G	R1,G	R2,G	W1,G	W2,G	G,G

 

 

The sample space for the first pull has five outcomes, {red, red, white, white, green}. Since the first pull is replaced, the sample space for the second pull is the same. For each choice of the first pull, there are 5 choices for the second pull. There are 5 • 5 or 25 possible outcomes:

 

Sample space: {(R1,R1), (R1,R2), (R1,W1), (R1,W2), (R1,G), (R2,R1), (R2,R2), (R2,W1), (R2,W2), (R2,G), (W1,R1), (W1,R2), (W1,W1), (W1,W2), (W1,G), (W2,R1), (W2,R2), (W2,W1), (W2,W2), (W2,G), (G,R1), (G,R2), (G,W1), (G,W2), (G,G)}

 

The event space for the first pull is the two red marbles. For each of those, there are two red marbles that can be chosen for the second pull. There are 2 • 2 or 4 outcomes in the event space:

 

Event space: {(R1,R1), (R1,R2), (R2,R1), (R2,R2)}

 

Again, notice that the size of the sample space for the two pulls is the product of the size of the sample space for each pull. Similarly, the size of the event space for the combined pulls is the product of the size of the event space for each pull.

 

Now, let's look at the probabilities for these situations, using the ratio of the size of the event space to the size of the sample space:

 

 

 

We can derive a formula from these data. Since the event space for a situation can be found by multiplying the event spaces of each independent event, and the sample space for the situation can be found by multiplying the sample spaces of each independent event, we have:

 

 

 

This is true for all situations with independent events. It can be extended to more than two events, as well.

 

If A and B are independent events, P(A and B) = P(A) • P(B).   In general, for any number of independent events, the probability that all the events happen is the product of the probabilities that the individual events happen.

 

Let's apply this to a problem:

 

 

 

Carlos has a deck of 15 cards numbered from 1 to 15. He pulls a card at random, looks at the number, and mixes it back into the deck. What is the probability that he does not get a 5 or less on the first try, but he does get 5 or less on the second try?   A)   B)   C)   D)   Show/Hide Answer A) Incorrect. You may have incorrectly defined the first event, or incorrectly found the probability for the first event. The correct answer is .   B) Correct. The first event (not getting a 5 or less) has probability  and the second event (getting a 5 or less) has probability . The events are independent because Carlos replaces the card each time, so the probability that he doesn’t get a 5 or less on the first card but does on the second card is .   C) Incorrect. You may have only found the probability for the second event and forgot to multiply it by the probability for the first event. The correct answer is .   D) Incorrect. You may have only found the probability for the first event and forgot to multiply it by the probability for the second event. The correct answer is .

 

 

 

Two (or more) events are independent if the occurrence of one event doesn’t change the probability that the other event occurs. There are two kinds of situations when this happens:

 

1) When the random action doesn’t remove an outcome (such as rolling a die or flipping a coin multiple times, or performing random actions that have no connection to each other like drawing a card and then rolling a die); and

 

2) when the random action does remove a possible outcome, but the outcome is replaced before the action happens again (such as pulling a card and then putting it back in the deck).

 

When events are independent, the probability that they all happen is equal to the product of the probabilities that each individual event happens.