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5.03 Probability of dependent events

Probability of dependent events

Two events are dependent if the outcome of one event affects the outcome of the other event.

For example, receiving a first place prize and receiving a second place prize in a race are dependent events. Once someone has won the first place prize, the sample space will decrease, so the probability of winning the second place prize will increase for those who are still in the race.

To find the probability of two dependent events, we multiply the probability of the first event by the probability of the second event.P\left(A\text{ and } B\right)=P\left(A\right)\cdot P\left(B\text{ after }A\right)

To understand this equation better, suppose you are selecting two cards from a deck of cards, one after the other. There are two ways this can happen:

  • Draw a card, then put it back in the deck before selecting another card

  • Draw a card and keep it, then select another card from the deck

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.

Exploration

To decide which chores Jacques needs to do, he pulls out pieces of paper from a hat. The options are sweeping (S), mopping (M), or vacuuming (V).

Tree diagram with first selection S M V; second selection M V for S; S V for M; and S M for V. Ask your teacher for more details.
Two chores on the same day
Tree diagram with first selection S M V and second selection S M V for each first selection. Ask your teacher for more details.
Two chores on different days
  1. Which tree diagram shows the experiment with replacement?

  2. What is the probability he will need to sweep and mop if they are done on the same day?

  3. What is the probability he will need to sweep and mop if they are done on different days?

  4. Is there a difference in the probabilities when selecting with or without replacement?

When selecting objects from a group "with replacement", selections are independent. Each time you select a card, you have the same probabilities.

This image shows a tree diagram for drawing two cards from a pack of 52 cards. Ask your teacher for more information.

For example, selecting a red card from a standard deck will be \dfrac{26}{52}=\dfrac{1}{2} every time if you replace the card.

This is because there will always be 26 red cards in the deck of 52 cards.

When selecting objects from a group "without replacement", selections are dependent. Each time you select a card, you change the probabilities for the next selection.

This image shows a tree diagram for drawing two cards from a pack of 52 cards. Ask your teacher for more information.

For example, the probability of drawing a red card from a standard deck is \dfrac{26}{52} = \dfrac{1}{2}.

If we draw a red card and do not replace it, there are only 25 red cards left and only 51 cards left in the deck. The probability of selecting a second red card is \dfrac{25}{51}. This can be seen in the top branches of the tree diagram.

This shows the probability of the second event is dependent on what color card was drawn first.

Examples

Example 1

Esther was given four animal crackers as a snack. She has a donkey \left(D\right), an elephant \left(E\right), a goat \left(G\right), and a hippo \left(H\right). She eats one cracker and then another.

a

Are the selections independent or dependent?

Worked Solution
Create a strategy

Consider whether Esther replaced the first animal cracker before selecting the second animal cracker.

If Esther replaced the first animal cracker, the outcomes for the first and second choice will be the same. If Esther did not replace the first animal cracker, the outcomes for the first and second choice will be different.

Apply the idea

Since the first cracker is eaten, it is not available in the second round. This means the selections are dependent events because the outcome of the second selection will depend on which animal cracker was eaten first.

b

List the sample space for the possible two crackers she eats.

Worked Solution
Create a strategy

We can make a list of all the outcomes or construct a tree diagram to visualize all the outcomes. Since the events are dependent, the number of options for the second cracker will reduce by 1.

Apply the idea
A tree diagram with Several directed edges connecting points labeled D, E, G, and H, with arrows pointing from one point to another and labels indicating the direction. Speak to your teacher for more information.

The sample space is DE, DG, DH, ED, EG, EH, GD, GE, GH, HD, HE, HG.

Reflect and check

For the first animal cracker, she has 4 choices. After eating one, she has 3 choices for the next one. By the fundamental counting priciple, there are 4 \cdot 3=12 outcomes in the sample space.

c

Find the probability that Esther eats the elephant first, then the goat.

Worked Solution
Create a strategy

We can use the tree diagram from part (a) to find the number of desired outcomes and the total number of outcomes in the sample space.

Apply the idea

From part (a), we found the total number of outcomes in the sample space is 12.

There is only one outcome where Esther eats the elephant first, then the goat \left(EG\right).

The probability is \dfrac{1}{12}.

Reflect and check

We can use the formula for dependent events to check our answer. There is only 1 elephant out of the 4 crackers, so the probability of eating the elephant first is P\left(E\right)=\dfrac{1}{4}Since Esther ate one, there are now only 3 crackers left, and 1 of them is the goat. The probability of eating the goat next isP\left(G\text{ after E}\right)=\dfrac{1}{3}The probability of eating the elephant then the goat is

\displaystyle P\left(E\text{ and } G\right)\displaystyle =\displaystyle P\left(E\right)\cdot P\left(G\text{ after }E\right)Formula for dependent events
\displaystyle =\displaystyle \dfrac{1}{4}\cdot \dfrac{1}{3}Substitute the probabilities
\displaystyle =\displaystyle \dfrac{1}{12}Multiply

Example 2

A pile of playing cards has 4 diamonds and 3 hearts.

a

Find the probability of selecting two hearts if two cards are selected from the pile with replacement.

Worked Solution
Create a strategy

To model the situation, a tree diagram can be drawn. The sample space is:

  • Diamond, diamond

  • Diamond, heart

  • Heart, diamond

  • Heart, heart

A graph consisting of square and circular nodes connected by lines. Each square node is labeled either Diamond or Heart, while the circular nodes are unlabeled. The nodes are arranged in two horizontal levels with squares at the top and circles at the bottom. Lines connect each square node to one or two circular nodes below, forming a network of relationships between different types of labeled and unlabeled nodes.

Then, we can use the diagram to find the probability of selecting a heart and another heart. If we let event A represent drawing a heart first and event B represent drawing a heart second, we can use the notation P\left(A\text{ and } B\right).

Apply the idea

For the first card, the probability of drawing a diamond is \dfrac{4}{7}, and the probability of drawing a heart is \dfrac{3}{7}. The probabilities for the second card are independent of which card was drawn first, so they will be the same.

The probability tree shows this situation with the correct probability on each branch:

A probability tree. First two branches are Diamond and Heart with probability of 4/7 and 3/7 respectively. Then branched out from these two branches are diamond and heart with the same probability.

Because these events are independent, we must use the formula {P\left(A\text{ and } B\right)=P\left(A\right)\cdot P\left(B\right)} to find the probability of drawing two hearts.

The probability of selecting a heart first is P\left(A\right)=\dfrac{3}{7}

The probability of selecting a heart, given a heart was selected first, is P\left(B\right)=\dfrac{3}{7}

The probability of selecting two hearts is

\displaystyle P\left(A\text{ and } B\right)\displaystyle =\displaystyle \dfrac{3}{7}\cdot \dfrac{3}{7}
\displaystyle =\displaystyle \dfrac{9}{49}
b

Find the probability of selecting two hearts if two cards are selected from the pile without replacement.

Worked Solution
Create a strategy

We can use the tree diagram of the sample space from part (a) but change the probabilities in the second set of branches since the first card is not replaced.

Apply the idea

For the first card, the probability of drawing a diamond is \dfrac{4}{7}, and the probability of drawing a heart is \dfrac{3}{7}. The probabilities for the second card are dependent on which card was drawn first.

The total number of cards will reduce to 6 because the first card was not replaced. If a diamond was selected first, then the number of diamond cards is now 3 and hearts will still be 3. If a heart was selected first, then the number of diamond cards is still 4 and hearts will now be 2.

The probability tree shows this situation with the correct probability on each branch:

A graph displaying relationships between nodes labeled Diamond and Heart at the top level, and unlabeled circular nodes at the bottom level. Speak to your teacher for more information.

Because these events are dependent, we must use the formula {P\left(A\text{ and } B\right)=P\left(A\right)\cdot P\left(B\text{ after } A\right)} to find the probability of drawing two hearts.

The probability of selecting a heart first is P\left(A\right)=\dfrac{3}{7}

The probability of selecting a heart given a heart was selected first is P\left(B\text{ after } A\right)=\dfrac{2}{6}

The probability of selecting two hearts is

\displaystyle P\left(A\text{ and } B\right)\displaystyle =\displaystyle \dfrac{3}{7}\cdot \dfrac{2}{6}
\displaystyle =\displaystyle \dfrac{6}{42}
\displaystyle =\displaystyle \dfrac{1}{7}
c

Is there a greater chance of selecting two hearts when the first card is replaced or when the first card is not replaced?

Worked Solution
Create a strategy

To find which situation gives us a greater chance of selecting two hearts, we can compare the probabilities we found in part (a) and part (b).

Apply the idea

When the first card is replaced, the probability of selecting two hearts is \dfrac{9}{49}.

When the first card is not replaced, the probability of selecting two hearts is \dfrac{1}{7}.

To compare, let's rewrite \dfrac{1}{7} as an equivalent fraction with a denominator of 49.\dfrac{1}{7}\cdot \dfrac{7}{7}=\dfrac{7}{49}Since \dfrac{7}{49}\lt \dfrac{9}{49}, there is a greater chance of selecting two hearts when the first card is replaced.

Example 3

Vanessa has 12 songs in a playlist. Four of the songs are her favorite. She selects shuffle and the songs start playing in random order. Shuffle ensures that each song is played once only until all songs in the playlist have been played. Find the probability that:

a

The first song is one of her favorites.

Worked Solution
Create a strategy

The probability can be calculated using P\left(\text{Event}\right)=\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}.

Apply the idea

Vanessa has 4 favorite songs and there are 12 songs in the playlist.

The probability that the first song is one of her favorites is \dfrac{4}{12}=\dfrac{1}{3}.

b

Two of her favorite songs are the first to be played.

Worked Solution
Create a strategy

The probability for the second song will be dependent on the first song that is played. We can calculate this probability using P\left(A\text{ and } B\right)=P\left(A\right)\cdot P\left(B\text{ after }A\right)where event A is a favorite song is played first and event B is a favorite song is played second.

Apply the idea

From part (a), we know the probability that the first song is one of Vanessa's favorites is \dfrac{1}{3}.

Because the songs cannot be repeated, this leaves 3 favorite songs out of 11 songs that have yet to play. So, the probability that the second song is also her favorite is \dfrac{3}{11}.

\displaystyle P\left(A\text{ and } B\right)\displaystyle =\displaystyle P\left(A\right)\cdot P\left(B\text{ after }A\right)Probability of dependent events
\displaystyle =\displaystyle \dfrac{1}{3}\cdot \dfrac{3}{11}Substitute known values
\displaystyle =\displaystyle \dfrac{1}{\cancel{3}}\cdot \dfrac{\cancel{3}}{11}Divide out common factor
\displaystyle =\displaystyle \dfrac{1}{11}Multiply
Idea summary

Two events are dependent if the outcome of one event affects the outcome of the other event. The probability of two dependent events is given by:P\left(A\text{ and } B\right)=P\left(A\right)\cdot P\left(B\text{ after }A\right)

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

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

Outcomes

8.PS.1

The student will use statistical investigation to determine the probability of independent and dependent events, including those in context.

8.PS.1a

Determine whether two events are independent or dependent and explain how replacement impacts the probability.

8.PS.1b

Compare and contrast the probability of independent and dependent events.

8.PS.1d

Determine the probability of two dependent events.

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