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10.02 Dependent events and replacement

Lesson

Dependent events

Two events are independent if the outcome of each event does not affect the outcome of the other event. The two events are not influence by each other. Two events that are not independent are called dependent.

If two events are independent then the following formula will be true: P(A \cap B) = P(A) \times P(B). Given data from an experiment we can check to see if this is significantly close.

Examples

Example 1

A standard six-sided die is rolled 691 times.

a

If it lands on a six 108 times, what is the probability that the next roll will land on a six?

Worked Solution
Create a strategy

We can use the formula: P(A)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}.

Apply the idea

The previous results do not affect the outcome of the next roll. There are 6 numbers on a die and 1 of them is a six.

\displaystyle P(A)\displaystyle =\displaystyle \dfrac{1}{6}Substitute the number of outcomes
b

Is the outcome of the next roll independent of or dependent on the outcomes of previous rolls?

Worked Solution
Apply the idea

The outcome of the next roll is not affected by the outcomes of previous rolls. So the outcome of the next roll is independent.

Idea summary

Two events are independent if the outcome of the each event does not affect the outcome of the other event. Two events that are not independent are called dependent.

To check if two events are independent you can use: P(A \cap B) = P(A) \times P(B)

Replacement

When selecting an card from a deck of cards, if we want to select a second card we have two choices:

  • Putting the card back in the deck before selecting another card

  • Keeping the card and selecting another card

The first method is described as "with replacement" because the card is placed back into the deck. The second method is called "without replacement" because the card is not placed back into the deck.

When selecting objects from a group "with replacement" selections are independent. Each time you select a marble or a card, you have the same probabilities each time. For example, selecting a red card will be \dfrac{26}{52}=\dfrac{1}{2} every time if you replace the card.

When selecting objects from a group "without replacement" selections are dependent. Each time you select a marble or a card, you change the probabilities for the next selection. For example, selecting a red card will be \dfrac{26}{52}=\dfrac{1}{2} the first time, however if you selected a red card on the first go, then the probability of selecting a red card on the second go is now \dfrac{25}{51}. There was one less red card to choose from and overall only 51 cards to pick from.

Examples

Example 2

A number game uses a basket with 10 balls, all labelled with numbers from 1 to 10. One ball is drawn at random, then another ball is drawn at random.

a

What is the probability that the ball labelled 2 is picked once if the balls are drawn with replacement?

Worked Solution
Create a strategy

Find the probability for each way that this could happen, then add them together.

Apply the idea

This could happen in two ways:

  1. P(2) \times P(2'): The ball labelled 2 is picked first and a ball not labelled 2 is picked second.

  2. P(2') \times P(2): A ball not labelled 2 is picked first and the ball labelled 2 is picked second.

\displaystyle P(2)\displaystyle =\displaystyle \dfrac{1}{10}Find the probability of picking the ball with 2
\displaystyle P(2')\displaystyle =\displaystyle \dfrac{19}{10}Find the probability of not picking the ball with 2
\displaystyle \text{Probability}\displaystyle =\displaystyle P(2) \times P(2') + P(2') \times P(2)Add the probabilities of the two ways
\displaystyle =\displaystyle \dfrac{1}{10}\times \dfrac{9}{10} + \dfrac{9}{10} \times \dfrac{1}{10}Substitute the probabilities
\displaystyle =\displaystyle \dfrac{9}{50}Evaluate and simplify
b

What is the probability that the ball labelled 2 is picked once if the balls are drawn without replacement?

Worked Solution
Create a strategy

Find the probability for each way that this could happen, then add them together. For the second event, the total number of balls will need to decrease.

Apply the idea

Similarly to part (a) this could happen in two ways. However, for the probability of the second ball, the total number of outcomes will be 10-1=9 since the ball was not replaced.

\displaystyle \text{Probability}\displaystyle =\displaystyle P(2) \times P(2') + P(2') \times P(2)Add the probabilities of the two ways
\displaystyle =\displaystyle \dfrac{1}{10}\times \dfrac{9}{9} + \dfrac{9}{10} \times \dfrac{1}{9}Substitute the probabilities
\displaystyle =\displaystyle \dfrac{1}{5}Evaluate and simplify
Idea summary

Events that occur "with replacement" have the item placed back into the group before each selection. Each selection is independent of the others.

Events that occur "without replacement" have the item remain outside of the group after selection. Each selection is dependent of the others. The probabilities will change each selection depending on previous selections.

Outcomes

VCMSP347

Describe the results of two- and three-step chance experiments, both with and without replacements, assign probabilities to outcomes and determine probabilities of events. Investigate the concept of independence.

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