Real easy probabily ( i forgot the basics)

[Neesan]

Carton of 10 Eggs, with 2 Bad ones. So we know the makeup of the population.

If we select 3 eggs at random.

What is the probability we have selected both bad eggs?

and

What is the probability we have selected at least one bad egg?.


Ive used tree diagrams and figured out the 1st question to be 1/15.






Im pretty sure there is more "numeric" way of calculating this as i find the tree process very tedious.

So if anyone could tell me how to do this using some formulas would be greatly appreciated.

 

[DaleSwanson]

As for part B the odds aren't 1/15. Just think about it for a second. I assume you know the odds of pulling a bad egg on a single draw are 1/5. How could the odds of pulling a bad egg over 3 draws be less?

What are the odds of pulling a good egg? If you know the odds of only pulling good eggs (on the three consecutive pulls), can you find the odds of pulling at least one bad egg using that?

 

[HallsofIvy]

To get two bad eggs and one good you can select in order (Bad, Bad, Good), (Bad, Good, Bad), or (Good, Bad, Bad).
The probability that the first egg selected is bad is 2/10= 1/5. There is now 1 bad egg and 8 good ones. The probability that the second egg selected is also bad is 1/9. After that the probabilty that the third egg selected is 1 (there are no bad eggs left. The probability of (Bad, Bad, Good) is (1/5)(1/9)= 1/45.

The probability that the first egg selected is bad is 2/10= 1/5. The probability that the second egg is good is then 8/9. There are now 1 bad egg and 7 good ones. The probability that the third egg is selected is bad 1/8. The probability of (Bad, Good, Bad) is (1/5)(8/9)(1/8)= 1/45.

The probability that the first egg selected is good is 8/10= 4/5. There are then 2 bad eggs and 7 good eggs left. The probability that the second egg is bad is 2/9. There are then 1 bad egg and 7 good eggs. The probability that the third egg is bad is 1/8. The probability of (Good,0 Bad, Bad) is (4/5)(2/9)(1/8)= 1/45.

The probability that one of those happens is the sum which is, of course, the 1/15 you give.

For "at least one bad egg", you have to add to that the probability of getting exactly one bad egg.

The probability of getting a bad egg on the first pick is, again, 2/10= 1/5. There is now one bad egg and 8 good. The probability that the second egg is good is 8/9. There is now one bad egg and 7 good. The probability of picking a good egg on the third selection is 7/8. The probability of (Bad, Good, Good) is (1/5)(8/9)(7/8)= 7/45. It is easy to show that the probability of each of the other orders, (Good, Bad, Good) and (Good, Good, Bad) is also 7/45 so the probability of "exactly one bad egg" is 3(7/45)= 7/15.

 

[LCKurtz]

Another way to look it is you have 8 good and 2 bad, so you want the number of ways you can choose 2 from the 2 bad and 1 from the 8 good divided by the number of ways you can choose 3 from 10:

$$P =\frac{\binom {2}{2}\binom 8 1}{\binom{10}{3}}$$

 

[CellCoree]

I would approach this problem using the principles of probability theory. In this case, we are dealing with a finite population of 10 eggs, with 2 of them being bad. Therefore, the probability of selecting a bad egg at random is 2/10 or 1/5.

For the first question, we are selecting 3 eggs at random. The probability of selecting both bad eggs would be (1/5)*(1/5)*(2/10) since we need to select the first bad egg, then the second bad egg, and then any of the remaining 8 good eggs. This simplifies to 1/125.

For the second question, we need to consider the probability of selecting at least one bad egg. This can be calculated by subtracting the probability of selecting all good eggs from 1. The probability of selecting all good eggs would be (8/10)*(7/9)*(6/8) since we need to select a good egg for each of the 3 selections. This simplifies to 14/45. Therefore, the probability of selecting at least one bad egg would be 1 - 14/45 = 31/45.

In summary, using the principles of probability theory, we can calculate the probability of selecting both bad eggs as 1/125 and the probability of selecting at least one bad egg as 31/45. This approach is more efficient and less tedious than using a tree diagram.