Let's see if you can solve this
Let's see if you can solve this
It's a 50% chance one is already heads, so it's basically just flipping a coin
1/3 easy
50%
Considering all options:
1.AA
2.AB
3.BA
4.BB
Let heads be A. This leaves us with just 1 and 2 left, of which 1/2 satisfies the requirement. Protip: this method works for all probability problems.
what a fucking idiot, lol. by guaranteeing one of the outcomes, it is no longer a variable. only one coin is a variable. 50%. also sequence doesnt matter, so there is no way to get 1/3. what a fucking moron.
>this method works for all probability problems
What about this one?
25% just like all the other outcomes
HH
HT
TH
TT
HT=TH
1/3
The way the question is worded you need to take conditional probability into account.
This really is a 50% chance.
the chances of grabbing a coin and flipping a head are 75% in this case. at the point of visible heads face up, it would 2/3 chance of all possibilities to be the trick coin. at the point of grabbing the coin, its 50% that she picked the trick coin.
what is an independent variable?
1/2 x 1/2 = 25% go get educated
1.A'
2.A'
3.A
4.B
A heads, ' marking the double headed. We eliminate 4 this leaves us with 2/3 satisfying the requirement. EZ.
Plus i did it before. The person is twice more likely to get heads on double-heads coin. This gives +1 to both alpha and omega which used to be accordingly 1 and 2. P=alpha over omega so 2/3
The chance of her taking out the coin is 50%
If it is a real coin, chance of a head is 50%
If it is the fake coin, it is 100% heads
Therefore, .5 x .5 x 1 = .25, 25%
Seriously, are all these people so dumb or just underage?
yeah, anyone that says 1/3 is pretty fucking dumb
Just look up the definition of probability, figure out the numbers and plug them in, duhh. Grade-school-level maths
Impossible because you're a loser and there's no way you can get head.
>mdw i realised i fucked up
I thought it was the first one heads, not at least one. Obv 1/3 mb m8s.