Question

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)

Answer 1

@butterflyprincess did you start working on the problem?

Answer 2

yea I gt 99/2500 for with replacement but I want to know if it is correct

Answer 3

ok, let me calculate and see if our answer matches

Answer 4

okay

Answer 5

My answer for with replacement is not matching with yours. Could you describe here how did you solve the problem?

Answer 6

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)

Answer 7

How A and B can occur at the same time?

Answer 8

oh no

Answer 9

P (A and B)= 0, so I think with replacement is solved now.

Answer 10

Hint: R1B2 = red first AND blue second

AND = multiply probabilities
OR = add them

Answer 11

wait so would it be then (32/1600)+(32/1600)

Answer 12

yep,

Answer 13

and then for without replacement it will be 32/1560 + 32/1560

Answer 14

because (4/40)(8/39)+(8/40)(4/39)

Answer 15

yes

Answer 16

yes, @butterflyprincess you got it right.

Answer 17

wait so if they ask P(R1 and G2) then it would be 0 since both events can't happen at the same time/?

Answer 18

P (R1 and G2)= probability of red bead first and green beard second is a valid event and can occur

Answer 19

which is 20/1600 (repl) and 20/1560 (without repl) right

Answer 20

but if they say P(R1 and G1) =0 since both red and green can't be first at the same time.

Answer 21

- but they cant occur together on one pick , of course

Answer 22

makes sense so that would be 20/1600 (repl) and 20/1560 (without repl) right?

Answer 23

correct :)

Answer 24

nice cool wait sorry I have another question lol
so I also need to find
P(No Yellows)
P(doubles)
and
P(no doubles)

Answer 25

for no yellows would it be 1089/1600 (with repl) and 1056/1560 (without)

Answer 26

yes, no yellows answer is right.

Answer 27

okay and how can I approach the doubles and no double problem

Answer 28

doubles mean.. same color both time.
So on both draws we should get same color

Answer 29

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)

Answer 30

cant you do like (1/6)(1/6) then without is (1/6)(1/5)

Answer 31

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.

Answer 32

do you understand?

Answer 33

yea oh okay so it would be like
(7/40)(7/40)) +(5/40)(5/40)......

Answer 34

right.. with replacement.

Answer 35

so would be like 0.1775 right right
then without replacement would be like
(7/40)(6/39)+(5/40)(4/39).....

Answer 36

yes, you are right.

Answer 37

probabil-ty of no doubles will i think be = 1 - p(doubles) right?

Answer 38

right! So like 1-0.1775!!

Answer 39

yes, p (non doubles)= 1 -p (doubles)

Answer 40

cool thanks for your help!
oh one more so P(G2 | R1) would be 5/40 (with repl) and 5/39 withut

Answer 41

Green in second provided Red is drawn first. Yes you are right.

Answer 42

nice thanks

Answer 43

no problem :)

Answer 44

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!!