anonymous
  • anonymous
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)
Mathematics
chestercat
  • chestercat
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ash2326
  • ash2326
@butterflyprincess did you start working on the problem?
anonymous
  • anonymous
yea I gt 99/2500 for with replacement but I want to know if it is correct
ash2326
  • ash2326
ok, let me calculate and see if our answer matches

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anonymous
  • anonymous
okay
ash2326
  • ash2326
My answer for with replacement is not matching with yours. Could you describe here how did you solve the problem?
anonymous
  • anonymous
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)
ash2326
  • ash2326
How A and B can occur at the same time?
anonymous
  • anonymous
oh no
ash2326
  • ash2326
P (A and B)= 0, so I think with replacement is solved now.
welshfella
  • welshfella
Hint: R1B2 = red first AND blue second AND = multiply probabilities OR = add them
anonymous
  • anonymous
wait so would it be then (32/1600)+(32/1600)
ash2326
  • ash2326
yep,
anonymous
  • anonymous
and then for without replacement it will be 32/1560 + 32/1560
anonymous
  • anonymous
because (4/40)(8/39)+(8/40)(4/39)
welshfella
  • welshfella
yes
ash2326
  • ash2326
yes, @butterflyprincess you got it right.
anonymous
  • anonymous
wait so if they ask P(R1 and G2) then it would be 0 since both events can't happen at the same time/?
ash2326
  • ash2326
P (R1 and G2)= probability of red bead first and green beard second is a valid event and can occur
anonymous
  • anonymous
which is 20/1600 (repl) and 20/1560 (without repl) right
ash2326
  • ash2326
but if they say P(R1 and G1) =0 since both red and green can't be first at the same time.
welshfella
  • welshfella
- but they cant occur together on one pick , of course
anonymous
  • anonymous
makes sense so that would be 20/1600 (repl) and 20/1560 (without repl) right?
ash2326
  • ash2326
correct :)
anonymous
  • anonymous
nice cool wait sorry I have another question lol so I also need to find P(No Yellows) P(doubles) and P(no doubles)
anonymous
  • anonymous
for no yellows would it be 1089/1600 (with repl) and 1056/1560 (without)
ash2326
  • ash2326
yes, no yellows answer is right.
anonymous
  • anonymous
okay and how can I approach the doubles and no double problem
ash2326
  • ash2326
doubles mean.. same color both time. So on both draws we should get same color
ash2326
  • ash2326
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)
anonymous
  • anonymous
cant you do like (1/6)(1/6) then without is (1/6)(1/5)
ash2326
  • ash2326
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.
ash2326
  • ash2326
do you understand?
anonymous
  • anonymous
yea oh okay so it would be like (7/40)(7/40)) +(5/40)(5/40)......
ash2326
  • ash2326
right.. with replacement.
anonymous
  • anonymous
so would be like 0.1775 right right then without replacement would be like (7/40)(6/39)+(5/40)(4/39).....
ash2326
  • ash2326
yes, you are right.
welshfella
  • welshfella
probabil-ty of no doubles will i think be = 1 - p(doubles) right?
anonymous
  • anonymous
right! So like 1-0.1775!!
ash2326
  • ash2326
yes, p (non doubles)= 1 -p (doubles)
anonymous
  • anonymous
cool thanks for your help! oh one more so P(G2 | R1) would be 5/40 (with repl) and 5/39 withut
ash2326
  • ash2326
Green in second provided Red is drawn first. Yes you are right.
anonymous
  • anonymous
nice thanks
ash2326
  • ash2326
no problem :)
ash2326
  • ash2326
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!!

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