Simple Probability: Understanding Odds of Picking Green Marbles from Bags

  • Thread starter Thread starter Paddy
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
In the discussion on simple probability, the first case involves a bag containing eight marbles, one green and seven red, resulting in a 12.5% chance (1 in 8) of picking the green marble. In the second case, with three identical bags, the probability of picking a green marble from each bag is calculated as 1/512, since the probability for each bag is multiplied together (1/8 * 1/8 * 1/8). The initial confusion arose from the assumption that the chances would be higher when considering multiple bags. The final clarification confirmed the calculations were correct, resolving the doubt. Understanding these probabilities is essential for accurately assessing odds in similar scenarios.
Paddy
Messages
24
Reaction score
0
I know this might be really simple, but I get easily confused.

CASE 1.

I have a bag with eight marbles, one of them is green and the remaining seven are red.

If you have to pick a random marble without looking, there is one in eight chances (12.5%) of lifting the green one.

Is this correct?

CASE 2.

I have three identical bags (like the one in case 1).

What are the chances of picking the three green ones? 3 in 24? (25%) this is how i though it would work out, but logically I predicted that it would have been more difficult to pick the three out of all the 24 (From three different bags), so I thought I was doing something wrong.

All help will be greatly appreciated.
 
Physics news on Phys.org
Hi There,

In terms of CASE 1:

P(green)=1/8

In Terms of CASE 2:

E1=result of the first bag
E2=result of the second bag
E3=result of the third bag

P(E1nE2nE3)=(1/8)*(1/8)*(1/8)=1/512

This should be correct, hope it helps

regards Steven
 
Last edited:
Thank you! That's what it was!

That does answer my doubt.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top