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 Math - Problem Set AA: Probability & Statistics, question #7
| (A) | 1/12 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 09/16/2012 22:03
Is there a faster way to do this than laying out the whole list? Thank you.
Posted: 09/19/2012 17:54
Jennifer, actually laying out the whole list is the fastest way for doing this, in my opinion. You can try and see how long it takes. In my case when I did it, it took less than a minute.
Posted: 01/02/2013 13:37
The probability of picking 3 from the jar is 1/4, without replacement the probability of picking zero on your second grab becomes 1/3, 1/4*1/3 = 1/12 for (3,0), repeat,
(0,3) = 1/4*1/3 = 1/12,
The other possible combinations is (2,1) and (1,2) each with a probability of 1/12,
1/12 + 1/12 + 1/12 + 1/12 = 4/12, reduce = 1/3
(0,3) = 1/4*1/3 = 1/12,
The other possible combinations is (2,1) and (1,2) each with a probability of 1/12,
1/12 + 1/12 + 1/12 + 1/12 = 4/12, reduce = 1/3
Posted: 04/12/2013 13:58
I did it a lot faster. I thought, okay, there are two combinations that give a sum of 3: 1+2 and 0+3.
So, I took each one separately. For 1 and 2: it doesn't matter which one I draw first. So the probability of drawing a 1 or a 2 is 2/4=1/2. Then there are 3 marbles left, and assuming I have one correct number, there is only one other correct number left, so the probability is 1/3. Multiply those and you have 1/6.
I started to do the same for 0 and 3 but quickly realized that it was the same patterns what I just did, just different numbers. So that gives another 1/6.
1/6 + 1/6 = 2/6 = 1/3.
So, I took each one separately. For 1 and 2: it doesn't matter which one I draw first. So the probability of drawing a 1 or a 2 is 2/4=1/2. Then there are 3 marbles left, and assuming I have one correct number, there is only one other correct number left, so the probability is 1/3. Multiply those and you have 1/6.
I started to do the same for 0 and 3 but quickly realized that it was the same patterns what I just did, just different numbers. So that gives another 1/6.
1/6 + 1/6 = 2/6 = 1/3.
Posted: 04/16/2013 19:20
Emily H, you figured out a different way to do it!
Posted: 07/29/2013 12:23
Why do you have to repeat the inversion? As in (3,0) and (0,3). If I pull out a 3 and then a 0, that's the same thing in the long run as pulling out a 0 and a 3, given that they add up to 3. There are really only two distinct ways of adding up to 3: (1,2) or (2,1), and (0,3) or (3,0) --as in, since we're adding the terms, the commutative property of addition should apply and should be that the order doesn't matter. That's why I'm arguing that there are only two unique ways to add up to 3. Therefore, it's 1/4*1/3= 1/12.
Can you clarify why pulling out a zero the. A 3 is different than pulling out a 3 then a zero?
Can you clarify why pulling out a zero the. A 3 is different than pulling out a 3 then a zero?
Posted: 07/30/2013 21:11
Yes, commutative property means 0+3 = 3+0. However, in statistics problem, drawing 0 then 3 and drawing 3 then 0 are two different things. It is not repeating, but counting properly different statistical events.
Posted: 01/22/2014 13:49
And what happen with (0,0)?
Posted: 01/31/2014 14:19
Gianpaolo, read carefully that the problem is WITHOUT REPLACEMENT. That means you cannot draw the same marble 2x.
Posted: 08/21/2015 08:22
I thought of it this way: After I pull the first marble, there are 3 marbles left. No matter which number I pull first, there is always exactly 1 marble, no more no less, that will lead me to add up to 3. So the answer is 1/3.