Consider a game that invovles two boxes, where inside each box there is a red ball and a blue ball.
The player first draws a ball from box 1 and place it inside box 2, then the player draws a ball from box 2.
If the ball being drawn out of box 2 has a different color than the ball being put in earlier, then the player wins.
- Let event E be the event where a red ball is drawn from box 1.
- Let event F be the event where a blue ball is drawn from box 2.
- Let event W be the event that the player wins the game.
Task:
- Calculate the probability of event F, P(F).
- Find out if event E and F are independent.
- Find out if event E and W are independent.
Task 1
This one's easy. I drew the probability tree for convenience.
P(F) = (2/6) + (1/6) = 1/2
Task 2
E and F are independent if: P(E|F) = P(E)
P(E|F) = P(E "INTERSECT" F)/P(F) = (1/4)/(1/2) = 1/2 = P(E)
So I guess they're independent? I wasn't expecting that.
Task 3
E and W are independent if: P(E|W) = P(W)
P(E|W) = P(E "INTERSECT" W)/P(W) = (1/6)/(1/3) = 1/3 = P(W)
Again, they're independent?
I find that hard to believe.
Surely event E (drawing a red ball from box 1) should have an affect on event F and W.
To say they're independent would mean that they don't affect each other at all. Is this possible?
Best Answer
The events $E$ and $F$ are independent if $\mathbb{P}(E \cap F) = \mathbb{P}(E) \cdot \mathbb{P}(F) $.
Using conditional probability, $\mathbb{P}(E \cap F) = \mathbb{P}(F |E)\cdot \mathbb{P}(E) = \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{6}.$
On the other hand, $\mathbb{P}(E) \cdot \mathbb{P}(F) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.$ Since $\mathbb{P}(E \cap F) \neq \mathbb{P}(E) \cdot \mathbb{P}(F)$ we conclude that the events $E$ and $F$ are dependent. The third task is similar.