If one egg is found to be good, then what is the probability that other is also good?

Solution 1:

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)

Solution 2:

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.

Solution 3:

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.

Solution 4:

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 $\dfrac12$

Solution 5:

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.