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This is a 9. grade elementary school quiz problem.

There is a bag with 4 marbles. Two of them are yellow and the other two are green. You blindly pick two marbles at once. What is the probability that you get two yellow marbles?

I would say the answer is $\frac{1}{6}$. There is a single combination to draw two yellow marbles and 6 possible combinations how to draw two marbles. My second approach is: possibility of picking the first one yellow is $\frac{1}{2}$, picking the second one yellow is $\frac{1}{3}$, multiplying possibilities gives me the same result $\frac{1}{6}$.

The problem is that my teacher claims that the answer is $\frac{1}{3}$. Her argument is that there are 3 possible outcomes - two yellows, two greens and a mixed one. One outcome is positive so therefore the answer is $\frac{1}{3}$. She also opposed to my second approach claiming that drawing marbles one by one is not the same as draw them simultaneously. From my opinion there should be no difference in result.

Could someone please write what is the correct answer to this problem? Does simultaneously drawing gives the same result as drawing one by one without replacement?

Finally, I think I understand the problem. Your answers help me to see it clearer. I found out that I mixed up the terminology. The problem has 6 outcomes and 3 possible events. The event that you draw the YG (mixed) combination has 4 positive outcomes while events of drawing YY or GG has a single possible outcome each. Probability is defined as a ratio between positive outcomes for specific event and all possible outcomes. Therefore the probability for drawing YY is $\frac{1}{6}$ and not $\frac{1}{3}$. By adding more green marbles you will increase the number of all possible outcomes, but the number of events will remain 3, so therefore the probability of picking YY will decrease and not stay $\frac{1}{3}$, what seems very logically. That was the catch I didn't see how to explain the teacher why I am correct. I think now I would be able to convince the teacher. Thank you all for your great help.

Matic Potocnik
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  • What is the sample space? – Andrew Chin Apr 11 '20 at 23:46
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    I suggest you read my answer here discussing why you might choose to treat flipping a head then a tail as a different outcome than a tail then a head when flipping two coins. – JMoravitz Apr 11 '20 at 23:48
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    The end result is that your teacher is wrong. Despite there being three discernable outcomes, this ignores several things, not least of which is the fact that you may only calculate probabilities using total number of "good" outcomes in the sample space divided by the total number of outcomes in the sample space in the event that those outcomes were equally likely to occur. You either win or lose the lottery, but winning the lottery doesn't happen $\frac{1}{2}$ of the time. – JMoravitz Apr 11 '20 at 23:49
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    If the marbles of the same colour are exactly the same, your teacher is correct: there are only three outcomes, YY, YG, GG ...and YG there is the same as GY as there is no order when drawing two marbles at once . – DonAntonio Apr 11 '20 at 23:50
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    @DonAntonio Being correct about the number of outcomes is one thing. Being correct about the probability of the event is another thing entirely. Don't add to the confusion by implying the teacher's answer for the probability was correct when it was not. – JMoravitz Apr 11 '20 at 23:51
  • For a more general formulation you may want to read about the hypergeometric distribution. –  Apr 11 '20 at 23:51
  • @JM In this case, outcomes is the same as sample space, of course. And yes: I think the teacher's answer was the correct one... – DonAntonio Apr 11 '20 at 23:53
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    @DonAntonio Take the situation with replacement. That's the same as flipping a fair coin twice. Once again there are exactly three "outcomes", namely a pair of Heads, a pair of Tails, or a mixed pair. Do you believe that the probability of each outcome is $\frac 13$? It's easy enough to try it... – lulu Apr 11 '20 at 23:59
  • @Lulu, that is crytal clear for mebecause it is not the same 1 in the first coin and 2 in the second, or 1 in the second and two in the first one. But that differentiation is what I think disappears when you draw (flip coins) two balls at once. I think I may need to think this over more deeply. Probability was never my greatest love...to say the least. yet this is a very basic case. I'm still convinced the teacher was right. My point is: both balls of the same coin are indescernible..and that's the gist of this all. – DonAntonio Apr 12 '20 at 00:10
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    @DonAntonio rephrasing lulu's suggestion... flip two identical fair coins simultaneously and forget which coin came from your left hand and which came from your right... Maybe put them inside an opaque cup, cover it with your hand, shake it around, and then release. Repeat this fifty or so times... Yes, you will have only three discernable outcomes. That is not in question. What is in question is the probability of each to occur. You will find the mixed case where one came up head and one came up tails as occurring more often. – JMoravitz Apr 12 '20 at 00:15
  • @DonAntonio now... imagine that someone drew on one of the coins with invisible ink... or recognize that the coins in fact have a serial number on them that you never noticed before. Recognize that the probability of getting the mixed case is the same before you had noticed and gained the ability to discern between the coins and thus between the $TH$ and $HT$ cases as after you gained that additional insight. Having gained the ability to notice more details does not suddenly change the probabilities... and so imagining that such detail was present, even if it is not, also shouldn't change it – JMoravitz Apr 12 '20 at 00:17
  • Thanks for the comments. I really need to dive into this more thoroughly as I still don{t see it clearly. – DonAntonio Apr 12 '20 at 00:18
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    @DonAntonio the same issue arises when you roll a pair of dice. You can get a "$4$" in two ways, the easy way, ${1,3}$ or the hard way ${2,2}$. Again, if you like, you can call those two outcomes but the mixed outcome, ${1,3}$, is twice as likely as the double. – lulu Apr 12 '20 at 00:35

4 Answers4

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You are correct. The problem with the teacher's method is that the various outcomes are not equi-probable.

Since the numbers are so small, we can write out all the possible, equiprobable, draw sequences (assuming we take all four in some order). They are: $$YYGG\quad YGYG\quad YGGY\quad GYYG\quad GYGY\quad GGYY$$

We see that exactly one way in six works.

Note that four out of six give mixed draws, demonstrating that the three cases your teacher points out are not equiprobable. Indeed, your method works for the mixed case as well: The first draw can be of either color and then the next gives a mixed draw with probabilty $\frac 23$.

lulu
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    I don't think this is correct as YG is the same as GY when chooseing two marbles at once. I also don't know why you have a series of 6 outcomes with 4 letters each... Thus, there are only three outcomes: YY. YG, GG – DonAntonio Apr 11 '20 at 23:51
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    @DonAntonio Again, those three outcomes are not equiprobable, I am introducing the order in order to obtain equiprobable sequences. Just as it is sometimes useful to consider all $36$ equiprobable rolls of a pair of dice even though the rolls $(1,2)$ and $(2,1)$ give you the "same" outcome. – lulu Apr 11 '20 at 23:53
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    @DonAntonio I will suggest you read my linked question above as well about sequences of coin flips. Alternatively, consider the problem where we have $99$ identical red balls in a bag and only one blue ball and we draw one ball from it at random. What is the probability of having drawn the blue ball? Certainly not $\frac{1}{2}$ under any reasonable interpretation of the problem, this despite there being only two discernable outcomes... having pulled a red ball, or having pulled a blue ball. – JMoravitz Apr 11 '20 at 23:54
  • @Jm Of course not 1/2, certainly. When drawing two balls I understand that someone, without seeing, drew both ball at once. Then, I think,it is not relevant to call the outcomes (you put it the name you want) YG and GY, adn they are the same since balls of the same color are identical. – DonAntonio Apr 11 '20 at 23:56
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    @DonAntonio and that is perfectly find and reasonable to treat them as the same if you so insist... but in doing so you sacrifice the ability to use counting techniques to find probabilities! It is because we like working with counting techniques and sample spaces where every outcome is equally likely to occur that we may imagine that the balls were all uniquely labeled (even if in reality they weren't) or that we may imagine that the balls came out in sequence... so that we can work in an equiprobable sample space. – JMoravitz Apr 11 '20 at 23:59
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    @DonAntonio if there are 2 yellow and 1000 green, the probability of taking 2 yellow is still $\frac{1}{3}$ according to Your logic – acat3 Apr 12 '20 at 00:14
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Let's label the marbles for the sake of this question as 1 and 2 for each color $$Y_1, Y_2, G_1, G_2$$ If you choose two marble at the same time, the possible combinations are as follows:$$(Y_1, Y_2), (Y_1, G_1),(Y_1, G_2), (Y_2, G_1), (Y_2, G_2), (G_1, G_2)$$

This means there are $6$ unique outcomes, so the probability of the outcome we are looking for $(Y_1,Y_2)=\frac{1}{6}$. In this case you are correct as your teacher is lumping the middle 4 outcomes together as one "type" of outcome which is incorrect.

I hope this helps!

swolo
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Well, your teacher is definitely wrong and here's why. You consider 6 cases each of which has the same probability. You could write numbers 1 and 2 on yellow and same on the green marbles and count all of the outcomes as if all of the marbles are different and we take them out one by one. You would then get 12 possible cases (4 for the 1st marble and 3 for the 2nd one), two of which fit your request - both of them are yellow (Y1, Y2 and Y2, Y1) so the probability is 1/6. She, however, considers 3 cases 2 of which are equal (obviously the probability to take out 2 yellows is the same as 2 greens), but the other one (marbles are different) is more probable as it can be achieved by not 2 as each of the previous ones (Y1, Y2 or Y2, Y1) but 8 ways: Y1G1, Y1G2, Y2G1, Y2G2,... well, I guess you can finish this yourself. I don't know about your country but we in Russia have a quite famous among mathematicians anecdote on exactly this theme: "-What is the probability to go out to the street and meet a dinosaur? -one half - either you meet it or you don't". While there can be a certain number of cases you always have to check that all of them are equally probable

Alexey Kubarev
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You're completely right which makes your teacher completely wrong. I understand her intuitive approach but your analysis on 1/2*1/3=1/6 should have signal her that she's off course.

Allow me to throw all possible permutations from the universe Y1,Y2,G1,G2 (assuming the order mattes) and show what is the probability that you get two yellow marbles?

in this case you'll have:

('Y1', 'Y2') Good
('Y1', 'G1')
('Y1', 'G2')
('Y2', 'Y1') Good
('Y2', 'G1')
('Y2', 'G2')
('G1', 'Y1')
('G1', 'Y2')
('G1', 'G2')
('G2', 'Y1')
('G2', 'Y2')
('G2', 'G1')

Permutation result: $\frac{2}{12} = \frac{1}{6}$.

If we use combination (as suggested by @lulu) you'll get the same result: $\frac{1}{6}$.

adhg
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