anonymous
  • anonymous
Gaming Engineer Charles has manipulated one dice so it always shows a six. He also has three standard dices that are very similar to the manipulated dice. Charles and Sarah are gonna make the throws each with three dices. Charles obviously using the fake dice plus two standard dices while Sarah, who do not know about the manipulated dice, randomly selects three of Charles's four dices. How great is the probability that Sarah and Charles gets: 1) two sixes? 2) at least two sixes?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
one throw each****
phi
  • phi
For Question 1 Charles gets 1 six automatically. so you need to find the probability of getting exactly 1 six when rolling two dice. If you make a table of outcomes, 10 out of 36 will have exactly 1 six. Or do 1/6 chance of a six on the first die times 5/6 of no six on the second die that is 5/36. now multiply by 2 because the 2nd die could be the die with the six. Either way, you get 10/36 or 5/18 for the chance that Charles rolls exactly 2 sixes when he rolls 1 "funny" die and 2 normal dice
anonymous
  • anonymous
If we take both questions for charles first. so we start off with charles. as you said, Charles have a fake dice so he only needs to roll 1 six of 2 dices\[P(1six2dice)=\frac{ 1*5+1*5 }{ 6^2 } =\frac{ 5 }{ 28 } = 28%\] Charles 2) atleast two 6 total number of turnout- number of turnouts without a 6 = \[6^{2}-5^{2}\] \[P(atleasttwo6)=\frac{ 6^{2}-5^{2} }{ 6^{2} } = \frac{ 11 }{ 36 } = 31%\] right?

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anonymous
  • anonymous
so if we take sarahs case, first of all she has to pick 3 dices among 4, the probability that she chooses the cheating dice is 1/4 and that she chooses the other 3 is 3/4. if she doesnt have the cheating dice her probability is 15/3^3 and if she has the cheating dice its 15/3^2 \[P(two6on3dice)=\frac{ 3 }{ 4 }*\frac{ 15 }{ 6^3 }+\frac{ 1 }{ 4 }*\frac{ 15 }{ 6^2 }=\frac{ 135/27 }{ 864/27 } =\frac{ 5 }{ 32 }\approx 16% \]
phi
  • phi
yes, on charles (but you have a typo: 5/28 where you mean 5/18)
anonymous
  • anonymous
and then atleast two sixes total amount of turn-outs - amount without 6 = 6^3-5^3 + 6^2-5^2 \[P(atleast2sixes)=\frac{ 6^{3}-5^{3} + 6^{2}-5^{2} }{ 6^{3} + 6^{2} } = \frac{ 51 }{ 126 } = 40%\]
phi
  • phi
For sarah, if she has the cheating die, then that part of the calculation should be 1/4 * same prob as charles, right ?
phi
  • phi
in other words, if sarah has the cheater + 2 normal dice that is the same as Charle's (but multiply by 1/4 )
anonymous
  • anonymous
so for sarah just to roll 2 sixes her probability is 1/4 * 5/18 if she has the cheating dice?
phi
  • phi
If she has 3 normal dice it will be 1/6 * 1/6* 5/6 multiply that by 3 because we can switch the order of which die is not the 6 then multiply all of that by 3/4 3/4 * 3 * 1/6 * 1/6 * 5/6 + 1/4 * 5/18
phi
  • phi
For Sarah, at least 2 sixes, with 3 normal dice, we know the chance for getting exactly 2 sixes. add in the chance of getting 3 sixes, which is 1/6*1/6*1/6 and we get 3/4 * (3*1/6*1/6*5/6 + 1/6*1/6*1/6) for the cheater, it's the same as Charles 11/36, so 1/4 * 11/36 add those for the final answer.
anonymous
  • anonymous
@phi if sarah have the three normal dices you said "3/4 * 3 * 1/6 * 1/6 * 5/6 + 1/4 * 5/18" why do we add 1/4*5/18 in this one?
anonymous
  • anonymous
@phi "chance of getting 3 sixes, which is 1/6*1/6*1/6 and we get 3/4 * (3*1/6*1/6*5/6 + 1/6*1/6*1/6)" why can we just add the chances of getting 3 sixes to that?
phi
  • phi
***if sarah have the three normal dices you said "3/4 * 3 * 1/6 * 1/6 * 5/6 + 1/4 * 5/18" why do we add 1/4*5/18 in this one? *** That is confusing, isn't it? It is describing the "normal" term, and then I added in the cheater term to show the full answer.
phi
  • phi
**why can we just add the chances of getting 3 sixes to that?*** if we have the prob of getting exactly 2 sixes, we can add the prob of getting 3 sixes to find the prob of getting 2 or 3 sixes.

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