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

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- ash2326

@butterflyprincess did you start working on the problem?

- anonymous

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

- ash2326

ok, let me calculate and see if our answer matches

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## More answers

- anonymous

okay

- ash2326

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

- 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

How A and B can occur at the same time?

- anonymous

oh no

- ash2326

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

- welshfella

Hint: R1B2 = red first AND blue second
AND = multiply probabilities
OR = add them

- anonymous

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

- ash2326

yep,

- anonymous

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

- anonymous

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

- welshfella

yes

- ash2326

yes, @butterflyprincess you got it right.

- 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

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

- anonymous

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

- ash2326

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

- welshfella

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

- anonymous

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

- ash2326

correct :)

- 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

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

- ash2326

yes, no yellows answer is right.

- anonymous

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

- ash2326

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

- 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

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

- 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

do you understand?

- anonymous

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

- ash2326

right.. with replacement.

- anonymous

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

- ash2326

yes, you are right.

- welshfella

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

- anonymous

right! So like 1-0.1775!!

- ash2326

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

- anonymous

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

- ash2326

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

- anonymous

nice thanks

- ash2326

no problem :)

- ash2326

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