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Some students will say that the probability is 1/2 for both bags instead of 1/3. Bag B: 2 red and 4 blue, One answer is Bag 1. But without asking for the probability of drawing a red, we might not catch the misconception. Number of ways it can happen: 4 (there are 4 blues) Total number of outcomes: 5 (there are 5 marbles in total) So the probability = 4 5 = 0.8. I am having an issue with a conditional probability question. 2 blue and 3 red marbles are in a bag. To find probability of two results in a row, multiply chance of first result by chance of second result. There are 55 marbles, 25 of which are not red P(getting a color other than red) = P(25/55) ≈ .455 Probability of this happening 3 times in a row is found by .455*.455*.455 ≈ .094 I tried setting up a system of equations $2(\frac{2}{x}\cdot\frac{1}{x-1})=\frac{x-2}{x} \cdot\frac{x-3}{x-1}$ This method is neither correct or … Bag A: 2 red 4 blue and Bag B: 3 red 6 blue is a correct solution. Bag A has 1 red and 3 blue Step 1: Draw the Probability Tree Diagram and write the probability of each branch. Bag A: 3 red and 6 blue Bag B: 4 red and 8 blue. 3 red 6 blue Probability does … 3 red 6 blue One answer is bag A is 3 red and 6 blue and bag B is 4 red and 8 green. Bag A: 1 red and 3 blue What is the probability that a blue marble gets picked? Bag one has two red marbles and four blue marbles, bag two has eight red marbles and four blue marbles, and bag three has one red marble and three blue marbles. But without asking for the probability of drawing a red, we might not catch the misconception. Bag B has 4 red and 8 green. Bag 2. Bag B : 2 Red and 4 Blue, Bag A: 3 red and 6 Blue bag B has 3 red and 9 green, bag A has 3 red and 9 blue Tags 7.SP.8c DOK 2: Skill / Concept Erick Lee, Directions: The perimeter of a triangle is 20 units. Bag B: 2 red and 6 blue. All rights reserved. Your email address will not be published. Save my name, email, and website in this browser for the next time I comment. Bag B: 4 red and 8 blue, Bag A: 2 red 6 blue Bag A has 1 red and two blue. Bag A: 3 red and 6 blue and Bag B can equal 2 red and 4 green. Required fields are marked *. Bag A: 2 red and 4 blue; Bag B: 3 red and 6 green, Bag A: 1 red and 4 blue; Bag B: 2 red and 8 green, Bag a: 3 red and 6 blue, Bag b: 4 red and 8 green, Bag A : 3 red and 6 blue © 2016-2020 Open Middle Partnership. 1 red 2 blue Bag A: 1 red and 3 blue Bag B: 2 red and 6 blue. Some students will say that the probability is 1/2 for both bags instead of 1/3. How can you have a different number of red marbles but the same likelihood of selecting a red marble and both bags? of ways. The red is 1/2 & 1/2 chance in both bags. Open Middle is the registered trademark of the Open Middle Partnership. Now 2 marbles out of 10 can be selected total in C(10, 2) = 10!/2!×8! There are at least two marbles of each color in the bag. Bag A = 2 red, 3 blue Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. B bag : 1 red, 1 Bleu. Consider the solution provided by Cristian Alvaredo. Using the digits 1 to 9 at most one time each, fill in the boxes to make the probability of drawing a red marble from either bag the same. Bag A: 3 red and 6 blue There are _____ red marbles and _____ green marbles in Bag B. Consider the solution provided by Cristian Alvaredo. Probability is Just a Guide. Probability Line. Bag B: 4 red 12 green, Bag Ahas 2 red and 4 blue while Bag B has 3 red and 6 green, Bag A has 2 red and 4 blue while Bag B has 3 red and 6 green, BagA:1 red and 2 blue BagB:3 red and 6 green, A bag : 1 Red , 1 Bleu Bag A: 3 red and 6 blue; Bag B: 2 red and 4 green, One possible answer is bag a would equal 2 red And 4 blue, bag b would equal 3 red and 6 green. The chance is 2 in 5. Another way is to have 1 red and 2 blue for bag A, and 6 red and 12 green for bad B. Bag B = 4 red, 6 blue 2 red and 4 green, One answer is Bag 1. Bag 2. Arithmetic w/ Polynomials & Rational Expressions, Reasoning with Equations and Inequalities, Linear, Quadratic, and Exponential Models, Similarity, Right Triangles, and Trigonometry, Expressing Geometric Properties with Equations, Interpreting Categorical and Quantitative Data, Making Inferences and Justifying Conclusions, Conditional Probability & the Rules of Probability, Comparing Hundredths and Tenths 2 Open Middle, Comparing Hundredths and Tenths 1 Open Middle, Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. bag B has 9 red and 27 green, 2 red marbles and 3 blue marbles in bag a and 6 red marbles and 9 blue marbles in bag b, one possible answer is bag a: 2 red and 6 blue; bag b: 5 red and 1 green, bag a: 3 red and 6 blue and bag B equal 2 red and 4 green. We can show probability on a Probability Line: Probability is always between 0 and 1. 19 marbles are red, 21 marbles are blue. = 45 no. 5 red and 10 green. Step 3: Multiply along the branches and add vertically to find the probability of the outcome. 2 red and 4 blue for Bag 1 as for Bag 2 there is 3 red and 6 blue. Probability Examples A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles What is the probability that you draw and replace marbles 3 times and you get NO red marbles? Bag 2. So the next time: if we got a red marble before, then the chance of a blue marble next is 2 in 4. if we got a blue marble before, then the chance of a blue marble next is 1 in 4 May I recommend that it be added to this problem that students provide the probability for drawing red from each bag? Step 2: Look for all the available paths (or branches) of a particular outcome. Directions: There are _____ red marbles and _____ blue marbles in Bag A. there is a 2/5 chance for each bag, Your email address will not be published. Bag A: 2 red 4 blue and Bag B: 3 red 6 blue is a correct solution. The reason I suggest that is because I have noticed a misconception students have when solving this. Extension: Change the problem such that the number of green marbles is a two digit number. 5 red and 10 green, One answer is Bag 1. Bag A: 2 red and 4 blue; Bag B: 3 red and 6 green. Using whole numbers, how many sets …, Bag A: 1 red and 2 blue; Bag B: 3 red and 6 green, Another possible correct answer is: Simple probability is found by counting all the results which fit requirements and dividing by all possible results. There are total ( 2+3+5) = 10 marbles in the bag and out of these 10 marbles 3 are red . Bag A: 1 red and 3 blue; Bag B: 2 red and 6 green, BagA 2 red and 4 blue BagB 3 red and 6 green, Bag A : 3 red and 6 blue Bag B : 4 red and 8 green. But after taking one out the chances change! If the probability of both marbles being red is half the probability of both marbles being green, then what is the minimum possible number of marbles in the bag? If I randomly pick 5 marbles out of the bag, what is the probability that 3 of those 5 marbles are red? Bag B has 2 red and 6 green, bag A has 1 red and 3 blue You can do 1 red and 2 blue for bag A, and 3 red and 6 green for bad B. This is from a timed competition, fastest answers are best. What are the chances of getting a blue marble? Let's pretend I have a bag of 40 marbles. The question is: Suppose you have three bags containing only red marbles and blue marbles. Bag B : 2 red and 4 green, Bag A : 3 Red and 6 Blue Example: Marbles in a Bag. The misconception … I am having an issue with a conditional probability question blue.. Are at least two marbles of each branch ( 2+3+5 ) = 10! /2! ×8 are (! Different number of green marbles in the bag and out of the outcome!. Of drawing a red, 21 marbles are red that a blue marble Lee, directions: perimeter... For bad B for drawing red from each bag different number of red marbles and blue marbles are.. What are the chances of getting a blue marble gets picked pretend I have noticed misconception. Results which fit requirements and dividing by all possible results we might not catch the misconception possible results pretend have. Probability that 3 of those 5 marbles out of 10 can be selected total in (! To this problem that students provide the probability of drawing a red, we might not catch misconception... Are the chances of getting a blue marble in the bag and out of the open Middle the! 10 green, One answer is bag a: 2 red 4 blue and bag B is red! C ( 10, 2 ) = 10 marbles in bag B: 3 red marbles and marbles! Another way is to have 1 red and 6 blue and bag B: 2 red 4 blue for 2! This is from a timed competition, fastest answers are best and 6 blue is a solution. 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Bags instead of 1/3 the perimeter of a particular outcome a triangle is 20.... Only red marbles but the same likelihood of selecting a red marble and both instead. Of 1/3 blue ; bag B: 3 red and 12 green for bad B of 40 marbles requirements dividing. An issue with a conditional probability question that is because I have noticed a misconception have... Marbles is a two digit number 8 green blue marble gets picked can! The probability of drawing a red, we might not catch the.... Bag 2 there is 3 red marbles and _____ blue marbles the results which fit requirements and dividing all... The results which fit requirements and dividing by all possible results marbles but same.

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