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A basket contains 10 eggs out of which 3 are rotten. Two eggs are taken out together at random. If one egg is found to be good, then what is the Probability that other is also good?


I applied conditional probability. It says that one of them is good, so the probability of the other one being good can be found in the 9 eggs left out of which 6 are good, so Probability = $6/9$

Am I right with my understanding?

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    $\begingroup$ Why would it be $6/10$ when you stated that $6$ out of the remaining $9$ eggs would be good? $\endgroup$ – heropup Dec 18 '16 at 3:18
  • $\begingroup$ Yes you have only 9 remaining so it should be 6/9. $\endgroup$ – Kanwaljit Singh Dec 18 '16 at 3:20
  • $\begingroup$ @heropup, Sorry, a typo :) $\endgroup$ – Jon Garrick Dec 18 '16 at 3:31
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    $\begingroup$ Depends on what the statement "one of them is found to be good means". How is it found? Do you check a specific one (the first, the leftmost, the darker in color, any specific selection) and you discover it to be good, or you are told that (at least) one of the two eggs is good? These different interpretations yield different results. Yet another result, if you randomly select one of the eggs to test it and you find it good. $\endgroup$ – Thanassis Dec 18 '16 at 7:21
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    $\begingroup$ You need to define "one egg is found to be good". Do that mean one at random is good and the other is not yet known or both are inspected and at least one is good. $\endgroup$ – paparazzo Dec 18 '16 at 17:15

11 Answers 11

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Am I right with my understanding?

You're overlooking a subtle point here, one that is often overlooked. You do get the right result, but due to a bit of luck.

The problem is you're overlooking a choice — in addition to having selected a pair of eggs from the basket, you are also choosing to assign the labels "one" and "other" to the two eggs.

Correctly modeling how that choice is made extremely important.

Fortunately, the problem surely means for the choice to be made in the easier-to-understand fashion: either one of the following two equivalent models was used

  • The labels were assigned randomly, with each of the two choices having equal probability
  • The eggs were selected not as a pair, but one at a time, and the first egg selected being called "one" and the second egg selected being called "other"

The reason that this is surely right is that the problem appears to indicate either of the following typical procedures, which directly correspond to the two bullet points above:

  • Take a pair of eggs from the basket
  • Pick one of the two eggs
  • Check whether the chosen egg is good or rotten
  • Guess whether the other is good or rotten

or

  • Take one egg from the basket
  • Check whether it is good or rotten
  • Take another egg from the basket
  • Guess whether it is good or rotten

An example of a different procedure requiring a different model, incidentally, could be described as follows:

  • Take a pair of eggs from the basket
  • Your colleague checks whether they are good or rotten
  • If both are rotten, things stop here. Otherwise, your colleague tells you one is good
  • You ask your colleague to point to a good egg.
  • You now guess whether the other egg is good or rotten.

(the actual problem being modeled doesn't have to include a colleague; that character can just be a fictional entity that enables the analysis to actually pick out a good egg)

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  • $\begingroup$ The question states that the eggs are taken together, so only one of your scenarios applies here. Still the question is ambiguous and there are different answers depending on the interpretation. $\endgroup$ – Thanassis Dec 18 '16 at 12:08
  • $\begingroup$ > Two eggs are taken out together at random. I am taking out together and not with succession . $\endgroup$ – Jon Garrick Dec 18 '16 at 13:21
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    $\begingroup$ I think I am following. In poker we call it a down card. The second card is unknown and random until it is known. I take "If one egg is found to be good" as a random inspection but I can see your later interpretation. I wish you would have run numbers for the later interpretation. $\endgroup$ – paparazzo Dec 18 '16 at 17:10
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    $\begingroup$ The distinction highlighted by Hurkyl is known as the "Boy or Girl paradox" - en.wikipedia.org/wiki/Boy_or_Girl_paradox. $\endgroup$ – Meni Rosenfeld Dec 19 '16 at 10:33
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After picking one good egg.

Rotten eggs = 3

Good ones = 6

Total = 9

Probability (second is also good) = $\frac{6}{9}$ = $\frac{2}{3}$

Edit -

This question has some assumptions also.

How is it found?

Both are picked together not in succession.

When you are picking two eggs together there are two ways to select eggs. Do you check a specific one (then how?) and you discover it to be good, or you are told that one of the two eggs is good? These different interpretations yield different results. So I think question should be more specific to get answer accurately.

See this link also.

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    $\begingroup$ What about the probability of getting one bad and then one good? $\endgroup$ – DJohnM Dec 18 '16 at 7:08
  • $\begingroup$ @DJohnM, irrelevant. The first one is good. The events are independent, and one of them already happened. Edit: Now I think I see what you meant; see my comment on the question. $\endgroup$ – Wildcard Dec 18 '16 at 11:14
  • $\begingroup$ (-1), that's not the correct answer. $\endgroup$ – Santiago Dec 18 '16 at 16:29
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    $\begingroup$ @Santiago ok. According to you what should be the answer? $\endgroup$ – Kanwaljit Singh Dec 18 '16 at 16:34
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    $\begingroup$ (-1) This is not the correct answer to the question. This is the probability of picking a good egg after first having picked a good egg. See also Hurkyl's excellent answer. $\endgroup$ – Servaes Dec 19 '16 at 2:04
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I would interpret the problem statement somewhat differently. The space of all possible results when drawing two eggs of ten is

$$\Omega = \{ \{ x, y \} \subseteq \{ e_1, \ldots, e_{10} \} : x \neq y \}$$

with $|\Omega| = \displaystyle\binom{10}{2} = 45$. Then, let

$$A = \{ \{ x, y \} \in \Omega : \text{both } x \text{ and } y \text{ are good} \}$$

be the set of results with both eggs good and let

$$B = \{ \{ x, y \} \in \Omega : \text{at least one of } x, y \text{ is good} \}$$

be the condition that we know holds. The probability in question is therefore

$$P(A|B) = \frac{|A \cap B|}{|B|} = \frac{\binom{7}{2}}{45-\binom{3}{2}} = \frac{21}{42} = \frac{1}{2}.$$

This interpretation assumes that we don't know which one of the eggs is the good one, we just know at least one of them is.

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  • $\begingroup$ I agree with you here. The condition that one of the eggs is good reduces the sample space eliminating the possibility that two bad eggs are chosen. $\endgroup$ – Laars Helenius Jan 2 '17 at 18:27
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Yet another interpretation...

You pick two eggs, and sit looking at them.

The possibilities are that you picked:

  1. A Good followed by another Good, with probability$$P_{GG}=\frac7{10}\times \frac69=\frac{7}{15}$$
  2. A Good followed by a Bad, with probability$$P_{GB}=\frac7{10}\times \frac39=\frac{7}{30}$$
  3. A Bad followed by a Good, with probability$$P_{BG}=\frac3{10}\times \frac79=\frac{7}{30}$$
  4. A Bad followed by another Bad with probability$$P_{BB}=\frac3{10}\times \frac29=\frac{2}{30}$$

Note that these four outcomes have probabilities totalling exactly $1$.

A tricorder check tells you that your pair is giving off Good Egg Smell; in other words, one egg, at least, is known to be good. Case #$4$ is eliminated.

Of the three remaining cases, both Good has the same probability, $\frac7{15}$, as mixed Good-Bad, $\frac7{30}+\frac7{30}$. The correct answer is $\dfrac12$

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  • $\begingroup$ howver, the question askes for the probability of good, given that the first egg is good, which is 6/9, or just 2/3, your reasoning is flawed in that it assumes bad then good is a possible outcome that fits into "good given good" $\endgroup$ – Alex Robinson Dec 18 '16 at 10:27
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    $\begingroup$ @Cursed1701 The question does not state that the first egg is good. It states that the eggs are taken out together and one of them is good. $\endgroup$ – Thanassis Dec 18 '16 at 12:09
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Let $E_k$ be the event that the $k$th egg is good.

$P[E_2|E_1 ] = {P[E_1 \cap E_2] \over P[E_1]}$.

$P[E_1] = {7 \over 10}$, $P[E_1 \cap E_2] = { \binom{7}{2}\over \binom{10}{2}} = {7 \over 15}$.

Hence $P[E_2|E_1 ] = {10 \over 15} = {2 \over 3}$.

Comment:

The question is ambiguous, I interpreted "one egg is found to be good" as meaning the first checked egg is good.

Another interpretation is to compute $P[N=2|N\ge 1]$ which is easily computed to be ${ 1\over 2}$ (where $N$ is the number of good eggs in the selection).

However, I think this interpretation is less likely (unless elaborated otherwise) because if we assert $N\ge 1$, it means both eggs must have been checked, in which case the probability is academic.

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  • $\begingroup$ This assume that the first egg is good. Why couldn't the second egg be the good one? $\endgroup$ – Laars Helenius Jan 2 '17 at 18:29
  • $\begingroup$ @LaarsHelenius: Two things: 1. You are free to add your own enlightened answer rather than systematically downvoting those that you don't like. 2. The order refers to the order in which you check the eggs. How can I check the second egg without checking the first egg first? If you have checked both, there is no meaningful probability to talk about. $\endgroup$ – copper.hat Jan 2 '17 at 23:30
  • $\begingroup$ You are only told that one of the two eggs selected is good, not that the first one selected is good. Your solution does not include the potential for selecting a bad egg first followed by a good egg. As for adding my answer, there is no sense in repeating an answer supplied by some one else. I have upvoted the ones I think to be correct and downvoted the ones I think to be wrong and given a reason why I think that. You have been incredibly helpful to me in the past when I have needed help, but this answer is wrong. $\endgroup$ – Laars Helenius Jan 2 '17 at 23:36
  • $\begingroup$ @LaarsHelenius: I have a different interpretation, which I have elaborated above. $\endgroup$ – copper.hat Jan 3 '17 at 0:22
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The tricky part of the problem (not sure if it's inadvertent or by design) is the statement:

one egg is found to be good.

How is it found to be good? The question states that the two eggs are chosen together, and one egg is found to be good. This might rule out all the interpretations that consider a first egg and a second egg, but it is still vague. Making it specific changes the nature of the problem and the final answer. I see at least three interpretations:

  • We choose one specific egg out of the two eggs and we find it good. For example, if the eggs are placed side by side, we choose the leftmost. Or we choose the one with the darker colour (assuming that colour is independent of the egg being rotten). Any way/rule to chose the egg that is decided a priori and is independent of the egg being rotten, will do. We test the chosen egg and we find it good. We are then asked what is the probability of the other egg being good.
  • We do not test any egg, but instead we are told that one of the two eggs is good. We are then asked what is the probability of the other egg being good.
  • We choose one of the two eggs at random, we test it and we find it good. We are then asked what is the probability of the other egg being good.

Using the first interpretation we can simply say that since one egg is tested and known to be good we can get it out of our pool of eggs and we are left with 9 remaining eggs 6 of which are good. Hence the probability that the non-tested egg is good is $\frac69 = \frac23$

The second interpretation is a different scenario. We do know that a specific egg is good. We rather know that one of the two chosen eggs is good. So we know that we have either of these 3 possibilities: GG, BG, GB. If we were to calculate the probabilities of these events, under no conditions, we would find: $$P(GG) = \frac7{10}\cdot \frac69 = \frac{42}{90}\\ P(BG) = \frac3{10}\cdot \frac79 = \frac{21}{90}\\ P(GB) = \frac7{10}\cdot \frac39 = \frac{21}{90}\\ $$

So we notice that $P(GG) = P(BG) +P(GB)$. So it is equally probable to get 2 good eggs out of 2 eggs and getting one good egg out of 2 eggs. Now if we include the condition that the pair of eggs cannot be BB (because we know at least one is good) then the two events (one G or both G) are still equiprobable, and since they are the only possible events, each have probability $\frac12$

This interpretation is a variation of the boy-girl paradox. If I had to guess, I'd say that the problem probably wants to describe this situation. It just doesn't do a very careful job with the description.

The third interpretation yields the same results as the first one, but it's good to keep it distinct since in slightly different scenarios it can yield different results.

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There are 45 possible combinations of pairs removed (9+8+7+6+5+4+3+2+1).

3 of these combinations contain two rotten eggs (2+1).

21 of those combinations contain two good eggs (6+5+4+3+2+1).

The remaining 21 combinations contain one of each.

If you discover one of the eggs is good, then you know that you aren't in a situation where both eggs are bad. Leaving 42 possible combinations, 21 one of which have two good eggs. Chances are $\dfrac{21}{42}$ = 50%.

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  • $\begingroup$ This doesn't take into account that if you happened to pick a pair of eggs with one good and one rotten, you're less likely to find a good egg than if you picked a pair of eggs where both are good. $\endgroup$ – Carmeister Dec 18 '16 at 8:00
  • $\begingroup$ @Carmeister Should it? As far as I understand, we don't pick one of the two eggs at random to check if it's good. We check if at least one of the eggs is good, which doesn't involve randomness. $\endgroup$ – Adayah Dec 18 '16 at 12:00
  • $\begingroup$ You should clarify for this analysis you only know one is good. You don't know which one. The two were picked at random and you have been told at least one is good. $\endgroup$ – paparazzo Dec 18 '16 at 17:25
  • $\begingroup$ There is some ambiguity in the problem as stated, but in my defense I would cite testmanship. The problem isn't very interesting when interpreted the alternate way. $\endgroup$ – aepryus Dec 19 '16 at 8:45
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An alternate interpretation of the problem: you have taken two of the 10 eggs, and you know at least one of them is good. This means you are in the situation where either exactly one egg is good ($3 \cdot 7 = 21$ possible selections meet this criteria) or both eggs are good (${7 \choose 2} = 21$ possible selections).

So given that at least one of your eggs is good, the probability is $1/2$ that in fact both eggs are good.

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Think about selecting the first egg. We could have either a fresh egg or a rotten one. The probability of selecting a fresh egg is $\frac{7}{10}.$

The probability of selecting a rotten egg is $\frac{3}{10}.$

Since we already know that the first chosen egg is a fresh one, selection of second egg again draws two choices to it. Second egg could be either a fresh egg or a rotten one. But this time the probability of selecting either a fresh egg will be different, since the first chosen egg is a fresh one. We have a total of $9$ eggs , such that $6$ are fresh while $3$ are rotten, left now. The probability of selecting a second fresh egg is $\frac69$ Therefore, the probability that the other egg is a fresh one when it is known that the first egg selected is fresh $=\frac 69$

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    $\begingroup$ How do you know that the first egg drawn is a good one? $\endgroup$ – DJohnM Dec 18 '16 at 7:03
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after one egg is taken out..

rotten left . 3 good left 6 total 9 .

ways of selecting. good one.. is 6^p^1 total .. 9^P^1 so answer is

6^p^1/9^p^1.= 6 /9= 2/3

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  • $\begingroup$ Why should the first egg be the good one? What if the second egg is the one that is good? $\endgroup$ – Laars Helenius Jan 2 '17 at 18:32
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I think it's $\frac{2}{7}$ if you pick them together at the same time and not put them back in the basket.

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    $\begingroup$ It wouldn't hurt if you gave more details about how you deduced this. Otherwise, anybody could come and claim that the answer is $12323/12876 \pi$. $\endgroup$ – Alex M. Dec 18 '16 at 18:09

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