## Quiz 3: Part B number 2? |
ruhmone |
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You have two dice. One die is painted red with one blue side. The other die is all red. You randomly take a die from your pocket and toss it. All five sides that are showing are red. What is the probability that the side that is facing downward (the side that you cannot see) is blue? Are we supposed to assume that the die has already been tossed ? or that we are tossing the die and we are trying to figure out what the probability of the die landing on the blue side facing down? is the blue side facing downward for sure regardless of the die that we randomly choose? can someone clarify... if it is, then is the answer 50%? Because you only account for the probability of you choosing a certain die? |

neelems: Feb. 5, 2015, 11:52 p.m. |
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the answer isn't 50%. the die you see, the one with all 5 red sides, could be either of the two die. technically the side facing down could be either red or blue depending on which die it is. |

patrickHals: Feb. 6, 2015, 12:27 a.m. |
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I think the answer should be one half. It is given that the five sides are red. So then the sixth side is either blue or it is red. You do not have to calculate the probability that all five sides are red because you already know that it is true. |

AZobi: Feb. 6, 2015, 4:42 p.m. |
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My answer is 1/2 for the same reason as Patrick's. Given that everything you see is red, the side that is facing down can be either blue or red. The chance that it is blue is therefore 1/2. |

weisbart: Feb. 6, 2015, 8:35 p.m. |
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Perhaps you guys should label the sides and then write down all the possibilities for the side that is hidden? |

AZobi: Feb. 6, 2015, 9:01 p.m. |
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Is 1/2 wrong because it doesn't take into account the possibility that the die tossed is the die that is entirely red. If it was the entirely red die, then the possibility that the hidden side being blue would be 0% |

weisbart: Feb. 7, 2015, 11:31 a.m. |
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$\dfrac{1}{2}$ is wrong because it is not taking into account that you have a lot of extra information. In particular, five of the six sides that are showing are red even though, if it were the die with the blue face, there would only be a one in six chance of this occurring. This make it much more likely that the downward facing side is actually red. \[\] I want you guys to play with this one for a while. This problem though is just like the one we did in class where you have two cards in your pocket, one where both sides are red and one where one side is red and the other is blue, and you randomly take one from your pocket and put it on the table. If the side showing is red, what was the probability it is the red card? Remember how we did that one? Can you apply the technique we used to answering the current problem? |

martha2A: Feb. 7, 2015, 3:36 p.m. |
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After labeling each side, I got 1/6 to be my answer because there are 6 different outcomes that could occur (if side R1 is the facing me then when I flip it over Ill get R6, if R2 is facing me then when I flip it over I get R4, etc.), but only one red side facing me will make it so that when I flip it over the blue side will show. Did anyone get the same thing? |

danthytran2A: Feb. 7, 2015, 4:46 p.m. |
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Would it be possible to do this? There are 6 ways for dice 1 to land without showing any blue, since I'll say it's the dice with no blue side. For dice 2, which I'll say is the one with 1 blue side, there is 1 way for it to land without showing any red at all - exactly face-down on it's blue side. So I have 1 way for it to land on it's blue side facing the floor, and 7 ways for it to land without showing blue...1/7. Am I missing something here? I don't know if I approached this right. |

AZobi: Feb. 7, 2015, 7:20 p.m. |
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Martha2A, when I re-did my work, I got a similar, yet different answer. If the red I see is R1, then flipping it over will give me R6. I did this 6 different ways: (R1 and R6, R2 and R4, R3 and R5, R4 and R2, R5 and R3, R6 and R1, and finally the side that gives us blue). My answer was therefore 1/7. I am not 100% confident with this answer though :( |

Alan_Mendoza: Feb. 7, 2015, 8:41 p.m. |
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I believe it is 1/7, as well. |

dana: Feb. 7, 2015, 8:50 p.m. |
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I don't know how you got 1/7. I got 1/12 because the other die has 6 possibilities as well. |

AZobi: Feb. 7, 2015, 9:54 p.m. |
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Wait, Dana, why would the other die have 6 possibilities? If you flip it over and it is anything but blue, doesn't that mean that the entire die is red? |

Alan_Mendoza: Feb. 7, 2015, 11:42 p.m. |
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dana, the the numerator is the number of ways the die has 5 visible red sides and 1 hidden blue side, that can only happen one way. The denominator is the number of ways the die can have 5 visible red sides. That's gonna be 6 from the all red die plus the 1 from the 5 red 1 blue die. There's only one possibility for the second die because the blue side has to be hidden in order for the remaining 5 sides to be red, the situations where the blue side is not hidden are not counted in the sample space.Therefore 1/7. Hopefully that clears it up. |

weisbart: Feb. 8, 2015, 10:09 a.m. |
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Very nice explanation, Alan_Mendoza. I hope this clears things up for everyone! |