1. ## seed germinating

the probability of a seed germinating is .6 , what is the probability that just one germinates

I thought the right answer would be .6*.4*.4*.4 but I have been told the right answer is .6*.4*.4*.4(4nCr3) i am not sure where this 4nCr3 comes from and i am not sure what its purpose/role is in the question

2. ## Re: seed germinating

.6*.4*.4*.4+.4*.6*.4*.4+.4*.4*.6*.4+.4*.4*.4*.6

You are running the experiment 4 times. You choose 3 of the experiments to not germinate. This leaves one experiment that will germinate. In general, if you run n experiments and you want the probability of exactly r successes, and the probabilty of success is p, the probability is:

$_{n}\text{C}_r p^r(1-p)^{n-r} = {_{n}\text{C}_{n-r}} p^r(1-p)^{n-r}$

(Note you can use either $_{n}\text{C}_r$ or $_{n}\text{C}_{n-r}$ because if you choose $r$ successes, then the remaining $n-r$ are automatically failures. If you choose $n-r$ failures, the remaining $r$ are automatically successes. So, you only need to do one, and they give the same answer.

3. ## Re: seed germinating

sorry I forgot to mention in the question that it is on a group of 4 seeds.
I understand this perfectly .6*.4*.4*.4+.4*.6*.4*.4+.4*.4*.6*.4+.4*.4*.4*.6
ok so it is to to do with the order, the first seed can germinate and the other 3 do not or the second seed can germinate and the other 3 do ETC. is this the right logic?

I am still kind of confused on the 4nCr3 . this tells us the different combinations that can be made from 4 objects taking 3 at a time. but in the question we are looking to arrange 4 objects not 3 ( to arrange the 4 seeds )

4. ## Re: seed germinating

Originally Posted by edwardkiely
sorry I forgot to mention in the question that it is on a group of 4 seeds.
I understand this perfectly .6*.4*.4*.4+.4*.6*.4*.4+.4*.4*.6*.4+.4*.4*.4*.6
ok so it is to to do with the order, the first seed can germinate and the other 3 do not or the second seed can germinate and the other 3 do ETC. is this the right logic?

I am still kind of confused on the 4nCr3 . this tells us the different combinations that can be made from 4 objects taking 3 at a time. but in the question we are looking to arrange 4 objects not 3 ( to arrange the 4 seeds )
Your 4 objects are the 4 trials. You needs to arrange three at a time that will not yield germinated seeds.

${_nC_r}$ has several meanings. It can also mean the number of ways to choose r items from a group of n items. You will sometimes see it written as (n choose r). So, you have 4 trials. Choose 3 that will not germinate. The number of choices is (4 choose 3) = ${_4C_3}$.

5. ## Re: seed germinating

.6*.4*.4*.4 is the probability that a specific one (the first when you have put them in some order) of four seeds germinates. .4*.6*.4*.4 is the probability that only the second one germinates. .4*.4*.6*.4 is the probability only the third one germinates, and .4*.4*.4*.6 is the probability only the fourth one germinates. The probability at one of the for germinates is .6*.4*.4*4+ .4*.6*.4*.4+ .4*.4*.6*.4+ .4*.4*.4*.6= 4(.6*.4*.4*.4)= [tex]_4C_1(.6*.4*.4*.4).