A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
You are choosing two beads
there are 40 beads.
7/40 are yellow
5/40 are green
9/40 are blue
8/40 are brown
7/40 are orange
4/40 are red
calculate WITH replacement and WITHOUT replacement for each:
P(R1B2 OR B1R2)
(R1 refers to chosing red the first time and B2 refers to picking Brown the second time)
anonymous
 one year ago
You are choosing two beads there are 40 beads. 7/40 are yellow 5/40 are green 9/40 are blue 8/40 are brown 7/40 are orange 4/40 are red calculate WITH replacement and WITHOUT replacement for each: P(R1B2 OR B1R2) (R1 refers to chosing red the first time and B2 refers to picking Brown the second time)

This Question is Closed

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2@butterflyprincess did you start working on the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea I gt 99/2500 for with replacement but I want to know if it is correct

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2ok, let me calculate and see if our answer matches

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2My answer for with replacement is not matching with yours. Could you describe here how did you solve the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yea so I used P(A or B)=P(A)+P(B)P(AandB) so I got like (32/1600)+(32/1600)(1024/2560000)

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2How A and B can occur at the same time?

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2P (A and B)= 0, so I think with replacement is solved now.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0Hint: R1B2 = red first AND blue second AND = multiply probabilities OR = add them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait so would it be then (32/1600)+(32/1600)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then for without replacement it will be 32/1560 + 32/1560

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because (4/40)(8/39)+(8/40)(4/39)

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2yes, @butterflyprincess you got it right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait so if they ask P(R1 and G2) then it would be 0 since both events can't happen at the same time/?

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2P (R1 and G2)= probability of red bead first and green beard second is a valid event and can occur

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is 20/1600 (repl) and 20/1560 (without repl) right

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2but if they say P(R1 and G1) =0 since both red and green can't be first at the same time.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0 but they cant occur together on one pick , of course

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0makes sense so that would be 20/1600 (repl) and 20/1560 (without repl) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nice cool wait sorry I have another question lol so I also need to find P(No Yellows) P(doubles) and P(no doubles)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for no yellows would it be 1089/1600 (with repl) and 1056/1560 (without)

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2yes, no yellows answer is right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay and how can I approach the doubles and no double problem

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2doubles mean.. same color both time. So on both draws we should get same color

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2either both yellows, or green or brown or blue or red or orange. find probability for each case and add them all P (Y1 Y2)+ P(G1 G2)+P(Br 1 Br 2)+P (BL1+BL2)+ P(R1 R2)+ P(O1 O2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cant you do like (1/6)(1/6) then without is (1/6)(1/5)

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2If we had only one bead for each color, we could do that. But the no. of each color beads is more than 1, we can't apply that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea oh okay so it would be like (7/40)(7/40)) +(5/40)(5/40)......

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2right.. with replacement.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would be like 0.1775 right right then without replacement would be like (7/40)(6/39)+(5/40)(4/39).....

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0probabilty of no doubles will i think be = 1  p(doubles) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right! So like 10.1775!!

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2yes, p (non doubles)= 1 p (doubles)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cool thanks for your help! oh one more so P(G2  R1) would be 5/40 (with repl) and 5/39 withut

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2Green in second provided Red is drawn first. Yes you are right.

ash2326
 one year ago
Best ResponseYou've already chosen the best response.2You can close the question, if you have no more parts of this question. Ask a new one if you need help with a different problem. Thanks!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.