Discussion
A bowl contains one marble labeled 0, one marble
labeled 1, one marble labeled 2, and one marble labeled 3. The bowl contains
no other objects. If two marbles are drawn randomly without replacement, what
is the probability that they will add up to 3?
*This question is included in Nova Press: Set V - Probability & Statistics, question #7
| (A) | 1/12 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 04/26/2012 02:28
I understand that 0+3 and 1+2 can get three. So why can't you do this
Prob of drawing a 0 Is 1/4 and prob of drawing a 3 is 1/3 so 1/3x1/4= 1/12
The same for the 1and 2 so 1/12 as well.
Adding 1/12+1/12=2/12=1/6
Prob of drawing a 0 Is 1/4 and prob of drawing a 3 is 1/3 so 1/3x1/4= 1/12
The same for the 1and 2 so 1/12 as well.
Adding 1/12+1/12=2/12=1/6
Posted: 04/30/2012 17:21
SM, I suppose you could.
In general, it's easier to list the permutations for a small universe problem like this. We have: (0,1), (0,2), (0,3), (1,0), (1,2), (1,3), (2,0), (2,1), (2,3), (3,0), (3,1), and (3,2). The pairs that have sum = 3 are (0,3), (1,2), (2,1), (3,0). That's 4 out of 12 possible combinations.
Your answer is wrong, because you need to add the combination of drawing 3 first, then 0. And add the combination of drawing 2 first then 1. then it's 1/12 + 1/12 + 1/12 + 1/12.
Remember, drawing 2 without replacement is the same as drawing 1 at a time until you draw 2, without replacement.
In general, it's easier to list the permutations for a small universe problem like this. We have: (0,1), (0,2), (0,3), (1,0), (1,2), (1,3), (2,0), (2,1), (2,3), (3,0), (3,1), and (3,2). The pairs that have sum = 3 are (0,3), (1,2), (2,1), (3,0). That's 4 out of 12 possible combinations.
Your answer is wrong, because you need to add the combination of drawing 3 first, then 0. And add the combination of drawing 2 first then 1. then it's 1/12 + 1/12 + 1/12 + 1/12.
Remember, drawing 2 without replacement is the same as drawing 1 at a time until you draw 2, without replacement.