Pulsified333
  • Pulsified333
Someone help I will give medal Question involving the rolling of two fair dice. What is the probability of one of the dice showing 3 or the sum being at least 8?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Well do you know how to compute probability?
Pulsified333
  • Pulsified333
yes I do, I thought the answer was 16/36 but it was wrong
anonymous
  • anonymous
So the probability of rolling 3 is equal on /both/ of them so we just leave that probability as 1/6th, not 1/12th. Does that help?

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anonymous
  • anonymous
Huh? On a fair dice there's a 1 in six chance of rolling a 3.
Pulsified333
  • Pulsified333
yes but can't you roll a 3 on lets say die 1 or die 2?
anonymous
  • anonymous
Yes, but that's 2/12 chance, which simplifies down to 1/6.
Pulsified333
  • Pulsified333
oh
anonymous
  • anonymous
Actually, 16/36 can be simplified. Maybe the answer needs to be simplified. What format is the test?
Pulsified333
  • Pulsified333
it simplifies it for you
anonymous
  • anonymous
Oh. well then.
anonymous
  • anonymous
So we've got 1/6th. Then we need to work out the chance of rolling a sum of of over eight and add it to 1/6th.
Pulsified333
  • Pulsified333
would it be over 36?
anonymous
  • anonymous
The denominator?
Pulsified333
  • Pulsified333
yes
anonymous
  • anonymous
IDK XD I'll work it out.
anonymous
  • anonymous
So how many combinations are there that equal above 8?
Pulsified333
  • Pulsified333
5
anonymous
  • anonymous
So there is a 5/36 (yeah, it is 36) chance of getting +8. So find a common denominator between 1/6th and 5/36ths and add them together.
Pulsified333
  • Pulsified333
i made a mistake
Pulsified333
  • Pulsified333
its the sum of at least 8
Pulsified333
  • Pulsified333
which equals 15/36
anonymous
  • anonymous
Ok. So still, multiply 'till you've got a common denominator + add them together.
Pulsified333
  • Pulsified333
quick question
anonymous
  • anonymous
Yeah?
Pulsified333
  • Pulsified333
wouldn't the probability of getting a 3 on one die be 10/36
anonymous
  • anonymous
I think the probability of 1 showing a 3 is 11/36. http://www.sms.rcs.k12.tn.us/TEACHERS/flaniganj/TEST/chapter_9_files/i0100000.jpg
anonymous
  • anonymous
Well 36/6 = 6 so it's 6/36
Pulsified333
  • Pulsified333
where is the 6 coming from?
Pulsified333
  • Pulsified333
would the answer be 25/36
anonymous
  • anonymous
You can get a 3 six different way on the first die, and 6 different ways on the second die. But two of those is (3, 3) so you subtract 1 to get 11/36 ways. 11/36 + 15/13 = 26/36
anonymous
  • anonymous
* 11/36 + 15/36 = 26/36
Pulsified333
  • Pulsified333
oh i see
Pulsified333
  • Pulsified333
the answer is still wrong
anonymous
  • anonymous
Well you're multiplying both sides of the fraction to give it 36 as a denominator, and you need to times 6 by 6 to get to 36, so so you also multiply 1 by six to get from 1/6 to 6/36, which you can add to 15/36.
anonymous
  • anonymous
Which by my admittedly thirteen-year-old brain is 21/36.
anonymous
  • anonymous
did you try the 25/36 you came up with? I'm guessing they want you to exclude the double roll.
Pulsified333
  • Pulsified333
yeah probably because it says one 3 not 2
anonymous
  • anonymous
was that it?
Pulsified333
  • Pulsified333
i didn't try it yet. I only have 2 attempts left
Pulsified333
  • Pulsified333
do you think 25/36 would be the right answer if its not including the double roll
anonymous
  • anonymous
yeah, not including it, it's 25/36. including it, it's 26/36 or 13/18.
Pulsified333
  • Pulsified333
no its not it either
Pulsified333
  • Pulsified333
;(
anonymous
  • anonymous
red = one 3 blue = sum at least 8|dw:1443039212787:dw|
anonymous
  • anonymous
there are 4 double counts. so they have to be subtracted. 10/36 + 15/36 - 4/36 = 21/36
anonymous
  • anonymous
I think

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