two marbles are selected at random from a box containing 5 red, 7 white and 8 blue marbles one after the other without replacement. find the probability that: a) both are red b) one is white and the other Is blue c) both are of the same colour.? Show
In Mathematics 4 Answers Available Asked by mercy on 18th February, 2020 Ask Your Own QuestionQuick QuestionsSee More Mathematics Questions
Related Pages What Is Probability Without Replacement Or Dependent Probability?In some experiments, the sample space may change for the different events. For example, a marble may be taken from a bag with 20 marbles and then a second marble is taken without replacing the first marble. The sample space for the second event is then 19 marbles instead of 20 marbles. This is called probability without replacement or dependent probability. We can use a tree diagram to help us find the probability without replacement. How To Find The Probability Without Replacement Or Dependent Probability?Step 1: Draw the Probability Tree Diagram and write the probability of each branch. (Remember that the objects are not replaced) Example: A jar consists of 21 sweets. 12 are green and 9 are blue. William picked two sweets at random. a) Draw a tree diagram to represent the experiment. b) Find the probability that i) both sweets are blue. ii) one sweet is blue and one sweet is green. c) William randomly took a third sweet. Find the probability that: i) all three sweets are green? ii) at least one of the sweet is blue? Solution: a) Although both sweets were taken together it is similar to picking one sweet and then the second sweet without replacing the first sweet.
Check that the probabilities in the last column add up to 1.
b) i) P(both sweets are blue) = P(B, B)
ii) P(one sweet is blue and one sweet is green) = P(G, B) or P(B, G) c) i) P(all three sweets are green) = P(G, G, G) ii) P(at least 1 sweet is blue) = 1 – P(all three sweets are green) What Is The Difference Between Probability With Replacement (Independent Events) And Probability Without Replacement (Dependent Events) And How To Use A Probability Tree Diagram?Examples:
Probability Question Using Tree Diagrams (Without Replacement)Example: A bag contains 5 blue balls and 4 red balls. A ball is picked and not replaced. What is the probability of picking at least one red ball?
How To Calculate Probability With And Without Replacement?A visual tutorial on how to calculate probability with and without replacement using marbles.
How To Calculate Probability Without Replacement Or Dependent Probability?Example: Andrea has 8 blue socks and 4 red socks in her drawer. She chooses one sock at random and puts it on. She then chooses another sock without looking. Find the probability of the following event P(red, then red).
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
There are 13 marbles, six green so 13- 6= 7 'not green'. So the probability the first marble is "not green" is 7/13. Then there are 12 marbles 6 "not green" so the probability the second is "not green" is 6/12= 1/2. The probability neither marble is green is (7/13)(1/2)= 7/26. The probability "at least one marble is green" is 1- 7/26= 26/26- 7/26= 19/26. You could also do this though it is harder: The probability the first marble is green is 6/13. There are then 12 marbles, 5 green, so 7 "not green". The probability the second marble is not green is 7/12. The probability of one green and one not green, in that order, is (6/13)(7/12)= (1/13)(7/2)= 7/26. The probability the first marble is not green is 7/13. There are then 12 marbles, 6 green so the probability the second marble is not green is 6/12= 1/2. The probability of one not green and one green, in that order, is (7/13)(1/2)= 7/26. Since "at least one green" includes "both green", the probability the first marble is green is 6/13. There are then 12 marbles, 5 green so the probability both marbles are green is 5/12 is (6/13)(5/12)= (1/13)(5/2)= 5/26. The probability of "at least one green marble" is 7/26+ 7/26+ 5/26= 19/26. Chris P. a. Both are red. b. The first marble drawn is red, and the second is blue. c. The first marble drawn is blue, and the second is red. d. One of the marbles is red and the other is blue. 3 Answers By Expert Tutors Probability is the # of possible outcomes desired / # of possible total outcomes. In problems like this when you have consecutive draws, you find the probability of each individual draw and then multiply your results. a. there are 4 reds to possible draw from the urn, there are 11 total marbles. The probability of drawing 1 red is 4/11. After the 4/11 chance that you did get a red, now what is the probability of getting a second red? *Note, the was no replacing so you have 1 red in your hand and 3 in the urn. Now there is a 3/10 probability of getting the second red marble. 10 because you have one of the 11 in your hand so there are only 10 left in the urn. so we have 4/11 and 3/10, multiply and we get 12/110 which reduces to 6 /55 b. same idea just keep your colors and number of marbles straight. P(b after a red has been drawn) = 2/10, 2 blue still in the urn but only 10 marbles since you have 1 red in your hand. Multiply 4/11 and 2/10 which equals 8/110 = 4/55 c. P(b) = 2/11, P(red after a blue has been drawn) = 4/10, multiply 8/110 = 4/55 d. this one is different, you have to account for either one being red or blue. so a red then blur or a blue then a red. We have already found these probabilities in b. and c. Since both situations need to be counted then we add them together. 4/55 +4/55 = 8/55 Why don't we multiply? Because that would give us a smaller probability and if you think about the problem, out chances increase if the order doesn't matter. (yes there is a mathematical explanation but I like to use common sense when possible)
Robert J. answered • 09/28/13 Certified High School AP Calculus and Physics Teacher
Andre W. answered • 09/28/13 Friendly tutor for ALL math and physics courses
Remember: P(A and B)=P(A)P(B) for independent events (which marble drawings are) P(A or B)=P(A)+P(B) for disjoint events 1. P(red and red) =(4/11)(3/10)=12/110 2. P(red and blue)=(4/11)(2/10)=8/110 3. P(blue and red)=(2/11)(4/10)=8/110 4. P(red or blue)=P(red and blue)+P(blue and red)=16/110 |