jestrs said:
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
cathcath said:
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.