Probability of eggs cracking homework

[jestrs]

1.A carton contains 12 eggs, 3 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what is the probability of the following events?

a. All of the cracked eggs are selected.
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected.



a. .0013
b. .0013
c. .0038

 

[EnumaElish]

Any ideas on how to tackle these?

 

[HallsofIvy]

1.A carton contains 12 eggs, 3 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what is the probability of the following events?

a. All of the cracked eggs are selected.
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected.



a. .0013
b. .0013
c. .0038

What, exactly, is your question? You give three problems and then three numbers. I can guess that your are saying that those are your answers and what to know they are correct. It would be far better if you were to show HOW you arrived at those answers.

 

[unplebeian]

If the a, b, and c are your answers they seem doubtful as the answers for a and b are the same. That means the probability of:

All of the cracked eggs are selected and
None of the cracked eggs are selected

are the same. This is very doubtful given the number of eggs. Also the probabilities seem quite low which means that it's next to nothing. The probability of selecting all 3 craked eggs SHOULD be next to nothing, but I'm trying to give you a feeling for the answer you have got. That's all.

 

[cathcath]

Having trouble as well

I am having trouble with this problem as well,
So far I am thinking P= #of possible combinations that include 3 cracked eggs/total #of 5 egg combinations possible.
For the denominator I get:
nCr= 12!/(12-5)!5!=12!/7!5!=792
I am stumped as to how to come up with the three different numerators.
I thought maybe the no cracked eggs might be 9!/(9-5)!5!=126??
Any help on how to set up the numerators for each of the 3 questions in the original post would be appreciated. Thanks

 

[unplebeian]

Look up Hypergeometric probability. That should give you a good hint.

 

[HallsofIvy]

1.A carton contains 12 eggs, 3 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what is the probability of the following events?

a. All of the cracked eggs are selected.
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected.



a. .0013
b. .0013
c. .0038

I am having trouble with this problem as well,
So far I am thinking P= #of possible combinations that include 3 cracked eggs/total #of 5 egg combinations possible.
For the denominator I get:
nCr= 12!/(12-5)!5!=12!/7!5!=792
I am stumped as to how to come up with the three different numerators.
I thought maybe the no cracked eggs might be 9!/(9-5)!5!=126??
Any help on how to set up the numerators for each of the 3 questions in the original post would be appreciated. Thanks

I wondered why a post from last December suddenly popped up again!

Just work it out one step at a time.

Initially you have 12 eggs , 3 cracked. The probability that the very first egg you choose is cracked is 3/12= 1/4, of course. If that happens then you have 11 eggs left, 2 of them cracked. The probability the second egg you choose is also cracked is 2/11. Assuming that happens you have 10 eggs left, one of them cracked. The probability the third egg is the one cracked one is 1/10. After that, of course, there are only uncracked eggs left, but to be complete, let's write those probabilities as 9/9 and 8/8 . The probability of "CCCUU" (first three eggs picked are cracked, last 2 are not) is (3/12)(2/11)(1/10)(9/9)(8/8).
But that's not the only way to get 3 cracked eggs. What about if the first you pick is uncracked, next three are. The probability the first egg is not cracked is (since there are 12- 3= 9 uncracked eggs) is 9/12. Now there are 11 eggs left and 3 are cracked: probability next is cracked is 3/11. Then 2/10, then 1/9 with the last uncracked of course, 8/8. Probability of "UCCCU" is (9/12)(3/11)(2/10)(1/9)(8/8). Notice that the numerator and denominators are exactly the same (with the numerators in different order)! That is exactly the same as the probability of "CCCUU". In fact, with a little thought, you should see that 3 "C"s and 2 "U"s in any order have exactly the same numerators and denominators, just with the numerators in different order. All you need to do is determine how many different ways you can write the letters "CCCUU"- and that's an easy formula- and multiply by the common probability.

I would think that "all uncracked" would be much easier: that's "UUUUU" and there is only one way to order that! Initially, there are 9 Uncracked eggs out of 12: 9/12; then there are 8 of 11: 8/11; then 7 of 10: 7/10, etc.

Do the last one, "two cracked, 3 uncracked", the same way.