20 May 16
You have a cubic die with 5 black faces and one white face.
You have a fair coin with one black face and one white face.
The coin is tossed and the die is rolled.
An adjudicator sounds a bell if he sees a white face.
You cannot see the coin or die ... but when you hear the bell ...
What is the probability that both faces are white?
1/2 or 1/6 or somewhere in between?
😀
21 May 16
Originally posted by wolfgang59
You have a cubic die with 5 black faces and one white face.
You have a fair coin with one black face and one white face.
The coin is tossed and the die is rolled.
An adjudicator sounds a bell if he sees a white face.
You cannot see the coin or die ... but when you hear the bell ...
What is the probability that both faces are white?
1/2 or 1/6 or somewhere in between?
😀
One in twelve. 1/2 x 1/6 = 1/12
21 May 16
Originally posted by wolfgang59
You have a cubic die with 5 black faces and one white face.
You have a fair coin with one black face and one white face.
The coin is tossed and the die is rolled.
An adjudicator sounds a bell if he sees a white face.
You cannot see the coin or die ... but when you hear the bell ...
What is the probability that both faces are white?
1/2 or 1/6 or somewhere in between?
😀
If you roll the die first, there's a 1 in 6 chance of it turning up white. After that you flip a coin which adds one more 'face chance' to the die roll (I hope that's right). So I'll say there's a 1 in 7 chance of a white face showing up in both the die roll and coin flip.
21 May 16
Originally posted by wolfgang59Is this a trick question?
You have a cubic die with 5 black faces and one white face.
You have a fair coin with one black face and one white face.
The coin is tossed and the die is rolled.
An adjudicator sounds a bell if he sees a white face.
You cannot see the coin or die ... but when you hear the bell ...
What is the probability that both faces are white?
1/2 or 1/6 or somewhere in between?
😀
If I cannot see the coin or die but I hear the bell, the probability that both faces are white is 100% (1/1)
21 May 16
3 edits
Originally posted by wolfgang59There's a 1 in 6 chance of getting white with the die. If both sides of the coin were white it would still be a 1 in 6 chance both die and coin show white. But the coin has 1 white side and 1 side black, so this adds one more number to the probability that both die and coin (together) will show white. With the die alone it's 1 in 6, with die and coin together it becomes 1 in 7.
Your first answer is correct but I'm not sure about your reasoning!
A coin flip that shows black doesn't add to the 1 in 6 because you're only looking for the probability of white showing up, so you're only adding the (one) side that shows white to the 1 in 6.
22 May 16
Originally posted by lemon limeIf you are throwing separately then hearing a bell tells you which throw resulted in the white face.
There's a 1 in 6 chance of getting white with the die. If both sides of the coin were white it would still be a 1 in 6 chance both die and coin show white. But the coin has 1 white side and 1 side black, so this adds one more number to the probability that both die and coin (together) will show white. With the die alone it's 1 in 6, with die and coin toge ...[text shortened]... bility of white showing up, so you're only adding the (one) side that shows white to the 1 in 6.
22 May 16
1 edit
Originally posted by sonhouseOh good grief, I assumed the adjudicator sounded a bell if he saw two white faces... but it says if he sees a white face. Apparently I got the right answer, but for the wrong reason.
If you are throwing separately then hearing a bell tells you which throw resulted in the white face.
The coin is tossed and the die is rolled.
An adjudicator sounds a bell if he sees a white face.
You cannot see the coin or die ... but when you hear the bell ...
What is the probability that both faces are white?
But after looking at this again, the problem doesn't seem to be much different from figuring out the probability of die and coin both showing a white face without an adjudicator and bell. If the coin is showing white there is a 1 in 6 chance the die will show white, and if the die is already showing white then there is a 1 in 2 chance the coin came up white... so it appears to be the same problem with or without the adjudicator.
I'll try to simplify my reasoning... when tossing the die there is a 1 in 6 chance of the white face showing. I think we can all agree with that. Adding the coin flip reduces the probability of both showing up white, but not by much. So it's simply a matter of figuring out by how much that 1 in 6 (die roll) probability has been reduced.
23 May 16
Originally posted by HandyAndyI commend your brevity!
There are 12 possible outcomes of the coin flip combined with the roll of the die
(WB WB WB WB WB WW and BB BB BB BB BB BW).
The probability of at least one white face showing is 7/12. The probability of one
of the seven showing two white faces is 1/7.
7/12 x 1/7 = 7/84 = 1/12
23 May 16
Originally posted by wolfgang59Please don't tell me that the probability of two white faces showing is zero.
I commend your brevity!
Please don't tell me that the adjudicator sounds a bell if he sees a white face
(i.e., one white face), and does not sound a bell if he sees two white faces.
Please don't tell me that! 😠ðŸ˜
23 May 16
Originally posted by wolfgang59[hidden]There are 7 possible bell ringers. Assuming #6 on the die is the white face of the die and the other five faces are black, we have the following bell ringers: 1W, 2W, 3W, 4W, 5W, 6W and 6B.
You have a cubic die with 5 black faces and one white face.
You have a fair coin with one black face and one white face.
The coin is tossed and the die is rolled.
An adjudicator sounds a bell if he sees a white face.
You cannot see the coin or die ... but when you hear the bell ...
What is the probability that both faces are white?
1/2 or 1/6 or somewhere in between?
😀
Of those 7, only one -- 6W -- is a double-white. So the odds of a bell-ringer being a double white are one out of seven.[hidden]