6th Grade - Independent And Dependent Events

 
     
 
     
 
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6th
Statistics
Independent and Dependent Events
Understand the difference between independent and dependent events.
Identify independent events and dependent events. Understand that the difference between independent and dependent events is that independent events don’t affect one another, while dependent events do. Know that independent events will always have an unchanging sample space (total number of possible outcomes), while dependent events will have an ever-changing sample space because of the affect of one event on the other. Use the sample space as the denominator of a ratio (fraction) to show a probability of an event, while using the possible favorable outcomes as the numerator (top number). Know how to find the probability of two independent or dependent events by multiplying the probabilities of the separate events. Calculate the probability of two independent events or dependent by multiplying the probabilities of each event.
 

Sample Problems

(1)

Identify which events are independent. Explain how you know.

  1. A bingo ball is removed from the machine, called out and placed to on the table. A second ball is removed.

  2. A bingo ball is removed from a machine, called out and placed back in the machine. A second ball is removed, called out and replaced.

(The events in B are independent because the original ball was replaced the player have the same number of chances of their number being picked as they did the first time. The sample space remains the same for all trials.)


(2)

Identify which events are dependent. Explain how you know.

  1. Heidi pulls a card from a deck, records the card number and puts the card in a new pile. She then pulls a second card, records it and places it in the new pile.

  2. Heidi pulls a card from a deck, records the card number and replaces the card at the bottom of the deck. She then shuffles the cards and pulls a second card.

(The events in A are dependent because the card from event 1 was never replaced. This changed the sample space for event 2.)

(3)

You are conducting an experiment. There are six cards with the numbers 1 – 6 on each card. What is the probability that you will roll a 4 on your first try and a 6 on your second try? Identify the kind of events described. Explain.

(1/36 or about 2.7%, independent events)

(4)

Tell whether the experiment described below is independent or dependent. Tell how you know.

A carnival game has 10 red cups, 25 yellow cups and 15 purple cups. The prizes awarded for red cups are stuffed animals, yellow cups, no prize and purple cups, a poster. You are give two ping pong balls to toss into the cups.

(Independent, the first event won’t affect the second event.)

(5)

Ian has a bag candy with; 5 red, 4 blue, and 1 green. He pulls a candy from the bag records the color than eats it. He then pulls another candy from the bag and repeats these steps.

Find P (red, green)

(5/90, 1/18, about 5.6%)

Learning Tips

(1)

Children need to understand the theoretical probability of an event occurring before they can solve to find independent and dependent probabilites. This section can be used to review this skill.


Probabilities range from 0 to 1, where 0 is impossible for an event to occur and 1 is 100% certain that an event will occur. Probabilities can be used to tell whether an event is; certain, very likely, equally likely, less likely and impossible to occur. Probabilities can be written as a fraction, a decimal or a percent. You may want to discuss how we hear probabilities in the media with weather forecasts, etc…


It is important that your child understand that when he/she finds theoretical probability, that he/she must always compare the number of favorable outcomes (probability being sought out) to the total number of possible outcomes (total possible combinations or choices).

The probability of an event is shown a P(n), where n is replaced with the event that is being solved for. So, if we were to be solving for the chances of spinning red on a spinner, we would write P(red) or P(r). Theoretical probability can be shown as a fraction. The numerator (top number) of the fraction will always be the number of favorable outcomes. The denominator (bottom number) of the fraction will always be the number of possible outcomes. The theoretical probability for any event would be shown as;


P(event) = number of favorable outcomes

number of all possible outcomes


The words would be replaced with numbers to solve for a specific probability. Here’s an example of how to use the formula above to find the theoretical probability of rolling an even number on a number cube 1-6.

  1. Write the probability you are seeking in the parenthesis after the P.

P (even)

  1. Count the possible favorable outcomes. The favorable outcomes for this problem are all even numbers on the cube. 2, 4, and 6 are even. Hence, there are 3 favorable outcomes of even numbers.

P(even) = 3__

  1. Count all possible outcomes. There are 6 numbers on the cube, which means a total of 6 possible outcomes.

P(even) = 3__

6

  1. If possible, write the fraction in simplest form by dividing both thnumerator and denominator by the greatest common divisor (largest number that will divide into both evenly).

So, 3/6 is the theoretical probability of rolling an even number.

(2)

It is important that your child know important probability vocabulary terms for this skill. Be sure that your child understands and can explain what each of the terms below means.

Sample space – the total number of items in the experiment, also known as the number of all possible outcomes. The sample space is always the denominator or bottom number of the ratio (fraction) used to show a probabilty. Ex. If a baggie has a total of 12 marbles, the sample space is 12.

Favorable outcomes – the number of possible specified outcomes. This means, if you’re asked to find the probability for red, you asked to find the favorable outcomes for red, or more simply how many red there are in all. The favorable outcome is always the numerator for the ratio used to show a probability.

(3)

It is important that your child be able to identify an independent event. An Independent event is an event that does not affect the next event. Independent events are only possible if an experiment’s sample space is unchanged. In other words, you must always replace the item drawn or removed in the first experiment before conducting the second experiment. This will mean that probabilities from independent events will always have the same denominator, or sample space. If the item is not replaced, the sample space will change and the outcome will be affected.


Independent Event Example: Tanya is drawing candies out of a bag of 12. She first gets a red, records her data and puts the candy back. On her second draw, she gets a green. This is an example of an independent event. Tanya replaced the candy before drawing a new one. This would give her the same denominator for each probability, 12. This is an independent event.


Be sure that your child understands that for an event to be independent, the sample space (denominator) must always be the same. This means the item must always be returned, or replaced, before the next event happens. You may want to review different terms used when an items is replaced. Here are a few;

  • Returned

  • Replaced

  • Put back

  • Gave back


(4)

Sixth graders need to be able to distinguish between independent and dependent events. It is important that they be able to identify a dependent event when they see one. Make sure your child understands that a dependent event is an event where the outcome of the first event affects the outcome of the second event. The reason the first event affects the second event is because the first event leaves the second event with a smaller sample size. A dependent event occurs when an item is removed from the sample space and never returned. Since the item is never returned, the sample space is one less than where it started from. This means, the items left in the second event will have a greater probability of being chosen.


Dependent Event Example: Tanya is drawing candies out of a bag of 12. She first gets a red, records her data, and then eats it. Next, she draws a green. This is a dependent event, because her first denominator or sample space would be 12, but after she ate the red, the sample space would go down to 11 for the next event. This would increase each colors chances of being chosen in the second event, because there are less to choose from.


Be sure that your child understands that for an event to be dependent, the sample space (denominator) must always be different from one event to the next. He/should understand that the sample space will become one less each time a new event occurs. For example, if the experiment above was completed 5 times, on the 6th event there would only be a sample space of 7 candies to choose from, because 5 out of the 12 would have been eaten in the other events.

(5)

Independent probabilities can be calculated by finding the product of the probabilities of both events. In order to do this, you will need to be able to first, determine the probability of each independent event. Next, you will need to multiply the two probabilities together. Here’s a step-by-step example:


Example: Monica has a bag of marbles. Inside the bag are 2 red, 4 blue, 5 orange and 1 black marble. What is the probability that she will reach into the bag and get a red marble, then a blue marble? Find the P (r, b) or P (red or blue).

1. Count the total number of marbles in the bag. This will be your sample space. 12 is the sample space. This will be the denominator for your fraction because it gives you the total possible outcomes.

2. Find the first probability, red. Look back at the original problem to determine how many marbles are red. Since 2 are red, there are two favorable outcomes for this event. This will be your numerator. Remember, the 12 for step one is your denominator.

2/12

3. Find the second probability. P (blue)

4/12

4. Multiply both probabilities together. P(red) x P(blue)

2/12 x 4/12 = 1/18

P (r, b) = 1/18 or about 5.5%


(6)

Similarly to finding an independent probability, dependent probability can be found by finding the product of the probabilities of both events. However, it is important to remember that this time each event will have a different sample space, or denominator. In order to do this, you will need to be able to first, determine the probability of each independent event. Next, you will need to multiply the two probabilities together. Here’s a step-by-step example:

Example: Mario has a bag of marbles. Inside the bag are 2 red, 4 blue, 5 orange and 1 black marble. Each time he reaches into the bag and pulls out a marble, he puts the marble on the ground. What is the probability that he will reach into the bag and get a red marble, then a blue marble? Find the P (r, b) or P (red or blue).


1. Count the total number of marbles in the bag for the first event. This will be your sample space. 12 is the sample space for the first event. This will be the denominator for your fraction because it gives you the total possible outcomes. ____

12

2. Find the first probability, red. Look back at the original problem to determine how many marbles are red. Since 2 are red, there are two favorable outcomes for this event. This will be your numerator. Remember, the 12 for step one is your denominator.

2_

12

3. Find the new sample space. Since Mario did not replace the marble, he has one less marble in the bag. 12 -1 = 11. The new sample space is 11. This will be used in the denominator of the second probability. ______

11

4. Find the second probability, using the new sample space. P (blue)

4_

11


5. Multiply both probabilities together. P(red) x P(blue)

2/12 x 4/11 = 2/33 or about 6%


Example 2: Mario has a bag of marbles. Inside the bag are 2 red, 4 blue, 5 orange and 1 black marble. Each time he reaches into the bag and pulls out a marble, he puts the marble on the ground. What is the probability that he will reach into the bag and get a red marble, then a blue marble? Find the P (b, b) or P (blue or blue).


To solve this problem, you would use the same steps as above, however, since you are looking for blue and already removed and placed a blue on the ground, the numerator on the second event will also become one less.


  1. Sample space 12

P(b) = 4, so the first probability is: 4_

12

  1. Sample space 11

3. The favorable outcomes for blue have changed, because only 3 blue marbles remain. (The other has been placed on the ground.) So, the second probability is: 3_

11

4. Multiply both probabilities together: P (blue 1) x P (blue 2)

4/12 x 3/11 = 4/44 or 1/11, about 9%


(7)

At times your child will be asked to write his/her probability as a percent. It is a good habit to be in to always do this. If your child needs a review on how to change a fraction into a percent, see below.


In order to turn a fraction into a percent, you will need to do two things.

  • Follow the steps to change the fraction into a decimal. (This step in simple terms is divide numerator by denominator). 3/12 = 3 12 = 0.25

Next, start at the decimal point and move it two places to the right. If the decimal stops behind the last number, you can just replace it with the percent sign. 0.25 = 25% However, if you have a decimal such as 0.375, you would keep the decimal and write 37.5%. Also, don’t forget, that if you have a decimal such as, 0.5 that you still need to move the decimal to places to the right. Since there is only one number to the right of the decimal, you add a 0 after the number. So, 0.5 is 50%.

Extra Help Problems

(1)

Identify the events as independent or dependent. Explain how you know.

Reggie is choosing items out of a grab bag. He chooses the first item, puts it in his pocket and reaches in for a second item.

(Dependent, the sample space was changed when Reggie stuck the first item in his pocket, instead of replacing it.)

(2)

Identify the events as independent or dependent. Explain how you know.

Stella has a bowl with 5 colors of sprinkles. She reaches into the bowl without looking and pulls out a sprinkle. She records the probability as 3/20. She reaches into the bowl again. This time she writes down 4/20.

(Independent, her denominator (sample space) did not change. This mean she must be putting the sprinkles back each time.)

(3)

Identify the events as independent or dependent. Explain how you know.

Daniel reaches into his sock drawer without looking and randomly grabs one sock and puts it on. He reaches in the drawer a second time and grabs another sock.

(Dependent, Daniel never returned the first sock to the drawer, so the sample space changed.)

(4)

Identify the events as independent or dependent. Explain how you know.

Chloe has a bag of multi-colored super balls. She reaches into the bag pulls one out and writes down the probability 1/5. She reaches into the bag second time, pulls out a ball and writes down 2/4.

(Dependent, the sample space was one less after the second event. This means Chloe is not replacing the balls in the bag after each event.)

(5)

Identify the events as independent or dependent. Explain how you know.

Michael has listed his first 5 probabilties from an experiment. They are:

1/24, 3/24, 2/24, 2/24, 7/24 and 3/24.

(Independent, the sample space is not changing.)

(6)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were to replace the item after the first event.

Find P(cube, pyramid).

(3/40, 7.5%)

(7)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were to replace the item after the first event.

Find P(sphere, rectangular prism).

(1/40, 2.5%)

(8)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were to replace the item after the first event.

Find P (cube, sphere).

(1/8, 12.5%)

(9)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were to replace the item after the first event.

Find P (pyramid or rectangular prism, cube).

(1/8, 12.5%)

(10)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were to replace the item after the first event.

Find P (cube, sphere or pyramid).

(1/5, 20%)

(11)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were not replacing the item after the first event.

Find P (sphere, cube).

(5/38, about 13%)

(12)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were not replacing the item after the first event.

Find P (rectangular prism, rectangular prism).

(2/380, 1/190, about 1/2% or 0.5%)

(13)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were not replacing the item after the first event.

Find P(cube, sphere).

(10/76, 5/38, about 13%)

(14)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were not replacing the item after the first event.

Find P (pyramid or rectangular prism, sphere).

(5/76, about 6.6%)

(15)

Suppose you put 5 spheres, 10 cubes, 3 pyramids and 2 rectangular prisms in a bag and removed them without looking. Find the probability if you were not replacing the item after the first event.

Find P(pyramid, pyramid).

(3/190, about 1.6%)

(16)

Suppose you have a box of tiles that contains; 2 red, 4 green, 3 blue, 2 orange, 1 purple. Without looking, you remove a tile, put it back and then select another.

Find P (o, g).

(1/18, about 5.6%)

(17)

Suppose you have a box of tiles that contains; 2 red, 4 green, 3 blue, 2 orange, 1 purple. Without looking, you remove a tile, put it back and then select another.

Find P (o, o).

(1/72, about 1.4%)

(18)

Suppose you have a box of tiles that contains; 2 red, 4 green, 3 blue, 2 orange, 1 purple. Without looking, you remove a tile, put it back and then select another.

Find P (r or g, b).

(1/8, 12,5%)


(19)

Suppose you have a box of tiles that contains; 2 red, 4 green, 3 blue, 2 orange, 1 purple. Without looking, you remove a tile, place it on the table and then select another.

Find P (b, b).

(3/66, about 4.5%)

(20)

Suppose you have a box of tiles that contains; 2 red, 4 green, 3 blue, 2 orange, 1 purple. Without looking, you remove a tile, place it on the table and then select another.

Find P (r or o, g).

(4/33, about 12.12%)

(21)

Suppose you have a box of tiles that contains; 2 red, 4 green, 3 blue, 2 orange, 1 purple. Without looking, you remove a tile, place it on the table and then select another.

Find P (r or o or p, g).

(5/33, about 15%


(22)

Lilliana has a bag of markers. In her bag are 2 red, 4 black, 3 pink, 2 brown, 1 green, 1 orange, and 1 yellow. If she pulls a marker from the bag without looking, replaces it and then pulls another marker out, what is the probability that she will first pull out a yellow or brown marker and then pull out a pink or red marker?

(15/196, about 7.7%)

(23)

Lilliana has a bag of markers. In her bag are 2 red, 4 black, 3 pink, 2 brown, 1 green, 1 orange, and 1 yellow. If allows Jason to take one marker without looking, keep it and then take a second marker, what is the probability that Jason will first pull out a green marker and then pull out a black or red marker?

(3/91, about 3%)

(24)

Johnny has a bag of candy. Inside his bag are 4 blue, 6 green, 5 brown, 2 red, and 3 orange candies. If Johnny reaches into the bag and grabs and candy without looking, eats it and grabs for a second candy, what is the probability that he will first get a blue or brown candy and then get an orange?

(27/380, about 7%)

(25)

Johnny’s brother has a bag of candy. Inside his bag are 2 blue, 8 green, 5 brown, 2 red, and 3 orange candies. If he reaches into the bag and grabs and candy without looking, puts it back in the bag, and then grabs for a second candy, what is the probability that he will first get a blue or green candy and then get a brown?

(1/8, 12.5%)

 

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