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@butterflyprincess did you start working on the problem?
yea I gt 99/2500 for with replacement but I want to know if it is correct
ok, let me calculate and see if our answer matches
My answer for with replacement is not matching with yours. Could you describe here how did you solve the problem?
oh yea so I used P(A or B)=P(A)+P(B)-P(AandB) so I got like (32/1600)+(32/1600)-(1024/2560000)
How A and B can occur at the same time?
P (A and B)= 0, so I think with replacement is solved now.
Hint: R1B2 = red first AND blue second AND = multiply probabilities OR = add them
wait so would it be then (32/1600)+(32/1600)
and then for without replacement it will be 32/1560 + 32/1560
yes, @butterflyprincess you got it right.
wait so if they ask P(R1 and G2) then it would be 0 since both events can't happen at the same time/?
P (R1 and G2)= probability of red bead first and green beard second is a valid event and can occur
which is 20/1600 (repl) and 20/1560 (without repl) right
but if they say P(R1 and G1) =0 since both red and green can't be first at the same time.
- but they cant occur together on one pick , of course
makes sense so that would be 20/1600 (repl) and 20/1560 (without repl) right?
nice cool wait sorry I have another question lol so I also need to find P(No Yellows) P(doubles) and P(no doubles)
for no yellows would it be 1089/1600 (with repl) and 1056/1560 (without)
yes, no yellows answer is right.
okay and how can I approach the doubles and no double problem
doubles mean.. same color both time. So on both draws we should get same color
either 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)
cant you do like (1/6)(1/6) then without is (1/6)(1/5)
If 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.
do you understand?
yea oh okay so it would be like (7/40)(7/40)) +(5/40)(5/40)......
right.. with replacement.
so would be like 0.1775 right right then without replacement would be like (7/40)(6/39)+(5/40)(4/39).....
yes, you are right.
probabil-ty of no doubles will i think be = 1 - p(doubles) right?
right! So like 1-0.1775!!
yes, p (non doubles)= 1 -p (doubles)
cool thanks for your help! oh one more so P(G2 | R1) would be 5/40 (with repl) and 5/39 withut
Green in second provided Red is drawn first. Yes you are right.
no problem :)
You 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!!