Michelle My Bell Posted November 25, 2015 Share Posted November 25, 2015 (edited) UPDATE: I posted another question below. I need help understanding how this problem is done... Suppose there are 9 blue marbles, 4 yellow marbles and 10 green marbles in a box. What is the probability of randomly selecting three green marbles with replacement? Show fractions (don't have to simplify final fraction) and then round to 4 decimal places. I don't even know where to begin. Edited November 25, 2015 by Michelle My Bell Quote Link to comment Share on other sites More sharing options...
Syllieann Posted November 25, 2015 Share Posted November 25, 2015 (edited) I assume "with replacement" means they put the marble they just drew back in before the next draw. If that is the case, then start with the fraction of marbles that are green (10/23). On each draw that is the chance of drawing green. There are three draws so (10/23) x (10/23) x (10/23) = your answer. You then need to do the decimal rounding business. Edited November 25, 2015 by Syllieann 3 Quote Link to comment Share on other sites More sharing options...
Michelle My Bell Posted November 25, 2015 Author Share Posted November 25, 2015 (edited) Awesome! That is exactly what I did after doing some research. Thank You! Edited November 25, 2015 by Michelle My Bell Quote Link to comment Share on other sites More sharing options...
Michelle My Bell Posted November 25, 2015 Author Share Posted November 25, 2015 OK, I have another one... My guesses are in Green and my thoughts are in Red. Assume that either Professor Givens or Professor Crowder has constructe the comprehensive exam that you must pass for graduation. There is a 60% chance that Givens wrote the exam and he asks a question about international relations 30% of the time. Crowder asks a similar question 75% of the time. If there is a question on the exam regarding international relations, what is the probability that Givens wrote the exam. A. Draw out and label the probability tree. Let I=International and N=Non-International. (At this point, I have concluded that P(I) is 30/60 for Givens and 75/40 for Crowder, AND P(N) is 70/60 for Givens and 25/40 for Crowder. I don't know how to construct the probability tree with this information. I can only come up with the following:) --------30/60 I --------70/60 N and ______70/60 I ______25/40 N (I don't know how to connect the two branches.) B.) What is the probability of getting a non-international question, given Professor Crowder constructed the exam? (I think maybe 25/40 = .625) C.) What is the probability of getting Professor Givens creating the exam and there is a question about International relations? (80/60 = 1.333) D. What is the probability of getting a non-international relations question on the comprehensive exam? (no idea) Quote Link to comment Share on other sites More sharing options...
silver Posted November 25, 2015 Share Posted November 25, 2015 (edited) Are you familiar with probability trees? It seems that part is maybe where you're getting stuck. A good thing to remember with probability is that a probability always has to be ≥ zero (never going to happen) and also ≤ 1 (always going to happen). So if you got an answer larger than 1, that means something wasn't calculated correctly. Edited November 25, 2015 by silver Quote Link to comment Share on other sites More sharing options...
Caroline Posted November 26, 2015 Share Posted November 26, 2015 (edited) It has been a while since I have studies this, but here is my attempt... Tree: -------------N .7 .6 G< --------------I .3 -----------N .25 .4 C< --------------I .75 So the probability of a non international question given Crowder wrote the test is P(N/C) = P(N intersect C)/P©=.4*.25/.25=.4 Probability of Givens creating exam and international relations is P(G intersect I)= .6*.3=.18 Probably of getting a non-international on the exam is P(C intersect N) + P(G intersect N)= .6*.7+.4*.25=.52 Edited November 26, 2015 by Caroline Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.