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:
$P(A\cap B)=P(A)\times P(B)$P(A∩B)=P(A)×P(B)
Given data from an experiment we can check to see if this is significantly close.
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:
$P(A\cap B)=P(A)\times P(B)$P(A∩B)=P(A)×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
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 $\frac{26}{52}=\frac{1}{2}$2652=12 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 $\frac{26}{52}=\frac{1}{2}$2652=12 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 $\frac{25}{51}$2551. There was one less red card to choose from and overall only $51$51 cards to pick from.
Worked exercise
Two cards are chosen at random from a deck of $52$52 cards without replacement. What is the probability of choosing 2 kings?
Think: The cards are drawn without replacement so the probabilities will change for each selection.
Do: On the first draw we have $52$52 cards, and there are $4$4 kings in the pack.
On the second draw, if we have already kept a king out - then we have $51$51 cards, and $3$3 kings still in the pack.
This results in the following probability of $2$2 kings being selected:
| $P\left(\text{3 Kings}\right)$P(3 Kings) | $=$= | $\frac{4}{52}\times\frac{3}{51}$452×351 |
| $=$= | $\frac{1}{221}$1221 |
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.
Practice questions
Question 1
A card is randomly selected from a normal deck of cards, and then returned to the deck. The deck is shuffled and another card is selected.
Are the events of each selection independent or dependent?
Dependent
AIndependent
B
Question 2
A standard six-sided die is rolled $691$691 times.
If it lands on a six $108$108 times, what is the probability that the next roll will land on a six?
Is the outcome of the next roll independent of or dependent on the outcomes of previous rolls?
Dependent
AIndependent
B
Question 3
A number game uses a basket with $10$10 balls, all labelled with numbers from $1$1 to $10$10.
One ball is drawn at random, then another ball is drawn at random.
What is the probability that the ball labelled $2$2 is picked once if the balls are drawn with replacement?
What is the probability that the ball labelled $2$2 is picked once if the balls are drawn without replacement?