A die is rolled twice. What is the probability of getting either a multiple of 3 on the first roll or a total of 8 for both rolls?

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A die is rolled twice. What is the probability of getting either a multiple of 3 on the first roll or a total of 8 for both rolls?

Mathematics
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This is what I have so far: P(multiple of 3)=2/6 P (sum of 8)=6/16 Then, you add both of them to get 31/48, but this is incorrect.
Wait I am not sure. Lemme think about it more
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Other answers:

Well, these are the possible answers: 5/12 7/18 13/180 4/9
Yea, I got 25/108 so something's off; I just don't know what?
Oh, I read the question wrong. I thought it was probability of of getting multiple of 3 and getting a roll of 8 on both rolls.
So, I keep getting different answers each time I complete the problem. Now I got 8/9 because I added 2/6 + 5/6=7/6 and subtracted that from 10/36 according to the Inclusive events formula P ( A or B)= P(A) +P(B)-P(A and B)
Its been a while. Gimme a few more minutes.
No, you're fine!
Got it. P(A)+P(B)- P( A and B)
For the multiple of 3 only three and six work.
2,6 3,5 4,4 5,3 6,2 All this would add up to 8. Now we would use independent event formula by multiplying probability by probability The chance of getting any of those number is
Equation bugged out.
Continued from 2 post above is \[\frac{ 5 }{ 6 }\] Chances for second dice would be 1/6 since you only need one number to make it 8. So \[\frac{ 5 }{ 6 } \times \frac{ 1 }{ 6 }=\frac{ 25 }{ }\]
5/36*
(5/36+2/6) -P(A and B) P(A and B) would be \[\frac{ 2 }{ 6 } \times \frac{ 1 }{ 6 }=2/36\]
You could have wait. If it was a timed test, you should have studied -_-
I understand how you have received all of these besides one part. (5/36+2/6) - P (A and B) I know that it's 2/6 from the multiple of 3 but where did you get 1/6? because I thought that P(total of 8 for both rolls)=5/36.
It wasn't a test, it's a lesson that I didn't fully grasp. I apologize for not responding sooner; I was working out the problem again. You've been wonderful help. Thank you!
Ah, I see. I was eating dinner, so sorry for the delay. P(A+B) is the probability for both event (A&B) to happen. A is the probability for it to be multiples of 3 which is 2/6 B is probability to add up to 8. If you rolled 3 on first dice, then second dice would need to be 5 to make it 8. If you rolled 6 on first dice, second dice would need to be 2 to make it 8. Because even if you were lucky to roll a 3 on first dice, you would need 5 on second dice. That would be a 1/6 chance so 2/6 x 1/6
Ah, you're wonderful at explaining things! Thank you so much!!!
Thanks for the feedback and nice job attempting it. Good luck on the rest!

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