Probability of Dependent Events
Just the opposite of independent events, dependent events are events in which previous attempts affect the outcome of subsequent events. Dependent events are just like they sound - each event is dependent upon what happened in the previous attempt. Let's look at an example of these dependent events.
James the Superb Magician likes to dazzle and amaze his audience with a card trick in which he selects two cards at random from a deck of cards but announces the cards that he will select prior to selecting them. James asks for an audience member to join him for the trick. Wendy, a female audience member, raises her hand to volunteer. The crowd gives her a rousing applause as she makes her way to the stage. James explains to Wendy that he will select a card at random from the deck of cards. Wendy examines the deck of cards to make sure the deck is fair. James announces that he will first draw an ace from the deck.
Wendy wonders: What is the probability that James will select an ace from the deck of cards? She knows that there are 4 aces out of 52 cards.
James explains to Wendy that he will now select another card from the deck, and it will also be an ace. Wendy is perplexed at how James could be lucky enough to draw two aces in a row from the deck of cards. She thinks to herself: What is the probability of James selecting an ace and then, without replacing the card, selecting another ace?
Wendy knows that to find the probability of him selecting an ace from the deck of cards, she will need to use the formula: total number of favorable outcomes over the total number of outcomes. In the first event, the number of favorable outcomes is 4 because James was selecting one of the 4 aces. The total number of outcomes is 52 because there are 52 cards in a standard deck of cards. So, Wendy knows the probability of James selecting an ace on the first draw is 4/52.
James turns over the first card that he selected to show it, in fact, was an ace. The crowd cheers with excitement.
Wendy now must calculate the probability of James getting an ace on the second draw. She knows that James did not replace the card, so there are only 51 cards left in the deck. She also knows that there could be only 3 aces left because the first card he selected was an ace. So, the probability of him getting an ace on the second draw is 3/51.
To find the probability of James getting an ace on the first card and then, without replacing it, getting an ace on the second card, Wendy needs to multiply these two events together. She will need to multiply 4/52 x 3/51. By multiplying these two probabilities together, she gets 12/2652. She reduces the fraction to 1/221. The probability of James selecting an ace from a deck of cards and then, without replacing the card, selecting another ace is 1/221.
Dependent Events Example
Let's look at another example: What is the probability of selecting a spade from a standard deck of cards and then, not replacing the card, selecting an ace? By looking at this question, we know that this is a dependent event because the card is not replaced. Therefore, the first event will affect the results of the second event. To calculate the probability of both of these events occurring, we will need to find the probability of each event separately and then multiply the two probabilities together.
To calculate the probability of the first event - selecting a spade from a standard deck of cards - we will use the formula: total number of favorable outcomes over the total number of outcomes. In a standard deck of cards, there are 4 different suits, each containing 13 cards. So, there are 13 spades in each deck of cards, which would be the number of favorable outcomes. The total number of outcomes would be the same as the number of cards in the deck, which is 52. The probability of selecting a spade from a standard deck of cards is 13/52. To make this problem easier, though, we can reduce the probability 13/52 to 1/4.
Now let's find the probability of the second event, selecting an ace from a standard deck of cards, occurring. Remember that after the first event, the card was not replaced. This affects the number of cards that are now in our deck of cards. With one card selected, there are only 51 cards remaining in our deck of cards. So, the total number of outcomes left is 51. As far as the number of favorable outcomes, one spade was selected in the first event. The card could have been the ace of spades. Since this card was discarded, the number of favorable outcomes that we know for certain remain in our deck of cards is 3. We know that the probability of the second event occurring is 3/51. Again, to make this problem easier, we can reduce this probability to 1/17.