Probability - Drawing balls from boxes

[Dell]

box A has place for 2 balls,
box B has place for 3 balls

balls are placed randomly in each of the boxes until one of them is full

if X is the amount of balls used, find:
E(x)
var(x)


i drew the boxes and think there are only 3 options,,, X=[2 3 4]

BOX A --- BOX B
I) 2 balls ---- 0 balls
II) 2 balls --- 1 ball
III) 2 balls --- 2 balls
IV) 1 ball --- 3 balls
V) 0 balls --- 3 balls

|__X || 2__|_3__|_4_|
|P(X) || 1/5 | 2/5 | 2/5|

E(X)=3.2

but the correct answer is 3.125

same problem (obviously) for var

 

[tiny-tim]

box A has place for 2 balls,
box B has place for 3 balls

balls are placed randomly in each of the boxes …

Hi Dell!

I think from your "5"s you're misinterpreting "randomly" …

you're treating it as 5 compartments, each equally likely,

but it's only 2 boxes, each equally likely.

 

[Dell]

not REALLY sure what you think i mean, but the 5 is not from 2+3 (balls) but rather the 5 options i mentioned before, - the 5 RANDOM ways the boxes can be filled
2-0
2-1
2-2
0-3
1-3

 

[tiny-tim]

ah … but those 5 ways are not equally likely.

Try again! ​

 

[Dell]

i thought that might be the problem, but can't fine the probability of each, since the likeliness of choosing either of the 2 boxes is 0.5, i thought the the probabillity of each possibility is 0.5^n (n being the number of balls used)

2-0 -- 1/4
2-1 -- 1/8
2-2 -- 1/16
0-3 -- 1/8
1-3 -- 1/16

but as you can see, this doesn't come to 1. so no point of even continuing

 

[tiny-tim]

ok, let's start by doing II) as an example (2A,1B) …

II) requires the third ball to be in A, and either the first to be in A and the second in B, or vice versa.

in short, II) is ABA +BAA, so there's two.

Carry on from there. ​