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Meehan98

  • one year ago

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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  1. Meehan98
    • one year ago
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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.

  2. anonymous
    • one year ago
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    Wait I am not sure. Lemme think about it more

  3. anonymous
    • one year ago
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    What is the correct answer?

  4. Meehan98
    • one year ago
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    Well, these are the possible answers: 5/12 7/18 13/180 4/9

  5. Meehan98
    • one year ago
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    Yea, I got 25/108 so something's off; I just don't know what?

  6. anonymous
    • one year ago
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    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.

  7. Meehan98
    • one year ago
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    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)

  8. anonymous
    • one year ago
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    Its been a while. Gimme a few more minutes.

  9. Meehan98
    • one year ago
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    No, you're fine!

  10. anonymous
    • one year ago
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    Got it. P(A)+P(B)- P( A and B)

  11. anonymous
    • one year ago
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    For the multiple of 3 only three and six work.

  12. anonymous
    • one year ago
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    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

  13. anonymous
    • one year ago
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    Equation bugged out.

  14. anonymous
    • one year ago
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    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 \[\frac{ 5 }{ 6 } \times \frac{ 1 }{ 6 }=\frac{ 25 }{ }\]

  15. anonymous
    • one year ago
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    5/36*

  16. anonymous
    • one year ago
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    (5/36+2/6) -P(A and B) P(A and B) would be \[\frac{ 2 }{ 6 } \times \frac{ 1 }{ 6 }=2/36\]

  17. anonymous
    • one year ago
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    You could have wait. If it was a timed test, you should have studied -_-

  18. Meehan98
    • one year ago
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    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.

  19. Meehan98
    • one year ago
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    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!

  20. anonymous
    • one year ago
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    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

  21. Meehan98
    • one year ago
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    Ah, you're wonderful at explaining things! Thank you so much!!!

  22. anonymous
    • one year ago
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    Thanks for the feedback and nice job attempting it. Good luck on the rest!

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