This is a new question, but related to this previous question:
http://answers.yahoo.com/question/index;_ylt=AhmYMgtYGCJqWzUALNxUmlgCxgt.?qid=20070127142145AAETN8z
The new question is this:
flip 2 (fair) coins and cover them both with my hand.
Then, I look at them both, tell you that at least one of them is heads, and take it out to show to you. The other is still covered, and you can't see it. Keeping in mind the answer to your the previous question, what are the odds that the covered coin is also heads?
2007-01-27 09:48:38 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Given that there are only 4 possible results, each with the same possibility:
H - H
H - T
T - H
T - T
If you show me that one of them is Heads, then the T-T possibility is gone. From the remaining 3 possible results, only ONE of them has the other coin being a heads.
Therefore, the answer is 1 out of 3, or 33%
2007-01-27 09:56:41 · answer #1 · answered by TankAnswer 4 · 1⤊ 0⤋
I like the two Qs, hit long rail and back, closest to the rail wins. I was always, almost touching it. EDIT-When I play, winner breaks, If you want to beak, beat me!, no letting the other guy break when I think he is a lesser player than I am, unless you are playing your buddies, and you know how they play, if you are playing for money, no player is going to show you what he's got, until the real money is put on the table. I played two guys once, they shot back to back, and I shot once, with the understanding that they cannot talk during the game, or they lose. They were too stupid to set each other up, that's why I played them in the first place. They lost 4 games, and the 5th, when one guy yelled at the other, set me up you f***.!, the bets and the Mattch ended 5-0 I won, and I dont even think it was anything special, I just new how stupid they were. It's like telling a guy I'll give you all the balls except one, and everybody that doesn't know takes that bet. The thing is if I cant run 8 or less balls, to beat him, I got no business playing for money, or spotting him the balls. The other guy is talking about a bar table, when I play on a regulation table, I don't like to a br, usually smaller table, but if I did it's all gravy for me, too easy!.
2016-03-29 05:26:26 · answer #2 · answered by Lorraine 4 · 0⤊ 0⤋
That's the same problem.
Conditional probabiity:
P( both heads | at least one head )=
P(both heads)/p(at least one)=
(1/4)/(3/4) = 1/3
Do you think that the problem is any different? I don't. It's the same conditional probability statement. I'd recommend that you learn to recognize conditional probability statements and how to solve them before you get yourself in trouble.
Btw, more than any other area of mathematics, more people will get problems wrong because they put too much weight on their intuition.
Oh, and with 3 doors, you double you chance of winning by switching doors once you're shown a door without the prize. Also a conditional probability question.
2007-01-27 10:08:24 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
1/2?
2007-01-27 10:14:07 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I agree with the guy above, it's 50-50.
2007-01-27 09:56:45 · answer #5 · answered by ♥Mi @m0r L@ UniC@♥ 3 · 0⤊ 0⤋
it's 50-50.
2007-01-27 09:52:40 · answer #6 · answered by muskrat_h8r 2 · 0⤊ 0⤋