Basic probability problem: Good Luck!

[DCal430]

I know I'm wrong but why isn't the second answer 28%?

1. Seeing Aardvark chance = 65%.
2. The chance it doesn't come with bear = 20%
3. .65 x .2 = 28%

Seeing a bear has 65% chance, seeing an ardvark has a 0.15/0.35 probability. Are some of you thinking that P(B) and P(A) are the same?

 

[DCal430]

I know I'm wrong but why isn't the second answer 28%?

1. Seeing Aardvark chance = 65%.
2. The chance it doesn't come with bear = 20%
3. .65 x .2 = 28%

FYI .65 *.20 = 0.13 which also incorrect.

 

[DCal430]

Let me break it down.

From

If you see an Aardvark you have a 65% chance of seeing a Bear.

We get P(Seeing Bear | Seeing Aardvark) = 0.65

From

You have an 80% chance of seeing either an Aardvark or a Bear

We Get P(Seeing Bear or Seeing Aardvark) = 0.80

From

Aardvarks do not lead to Bears and Bears do not lead to Aardvarks.

We Get P(Seeing Bear | Seeing Aardvark) = P(Seeing Bear)
P(Seeing Bear) = 0.65

Everything else should follow from this.

 

[mjrpes3]

So much for basic probability problem

I can't figure out how to visualize this problem in my head. I imagine the "seeing A then seeing B" as describing a series of combinations possibilities

single possibilities (one roll of dice)

A = Aardvark, B = Bear, 0 = None
A | B | 0 = 100%
A | B = 80%
0 = 20%
B = 65%
A = 15%

combination possibilities (two rolls of dice)

{AB,BA,AA,BB,A0,0A,B0,0B,00} = all combination possibilities
{AB,BA} = 2(.65 * .15) = ~20%
{A0,0A} = 2(.15 * .2) = 6%
{B0,0B} = 2(.65 * .2) = 26%
{00} = (.2 * .2) = 4%
{AA} = .15 * .15 = ~2%
{BB} = .65 * .65 = ~42%

This logic works but is not how DCal is seeing it. If I could visualize the basic premise I'm sure I could follow DCal's logic.

 

[Dr. Zaus]

Let me break it down.

From

We get P(Seeing Bear | Seeing Aardvark) = 0.65

From

We Get P(Seeing Bear or Seeing Aardvark) = 0.80

From

We Get P(Seeing Bear | Seeing Aardvark) = P(Seeing Bear)
P(Seeing Bear) = 0.65

Everything else should follow from this.

I just don't get why there's temporality between seeing A and then B...

But i'll play your game mr. Bond.

 

[Pia]

I just don't get why there's temporality between seeing A and then B...

But i'll play your game mr. Bond.

In the exercise we were given
P(B after A has already happened) = 0.65

In this particular case it's key to notice there isn't a temporality between B and A since they were told to be independent. A having happened has no effect on B happening. So the P(B|A) above and P(B) must be equal, and thus we obtain P(B)=0.65

 

[actuarial]

50/50.

Either it will or it won't.

That is the answer to any "probability" problem.

Thank god. The actuaries are wrong (they always are, just this time more so than usual).

Using this logic, we have an equal chance of dieing at each possible age, giving us a life expectancy of about 60 (omega of 120).

Apply this to the SS model and voila, it's fully funded.

This man just solved a good chunk of the debt crisis!

 

[nanette1985]

Apply this to the SS model and voila, it's fully funded.

This man just solved a good chunk of the debt crisis!

He did a better job of it than the people who are supposed to be in charge

 

[JTsyo]

50/50.

Either it will or it won't.

That is the answer to any "probability" problem.

If there are 3 choices do each of them have a 50% chance of being right? This will make multiple choice tests so much easier, thanks. :whiste: