Should I take counting arrangements into account with this question on probability?

Question

The probability a seed germinates is p and seeds germinate independently of each other. If n rows of m seeds are planted, find the probability there are 2 rows each containing 1 seed that fails to germinate, all other seeds germinating.
I got mp^3(n-2)(m-1)^2(1-p)^2.... do you think this is right?!

Answer 1

No correct answer yet.
One answerer is correct in that we need to determine the probability of 2 failing seeds being in two different rows.

The probability that two seeds fail and (nm-2) seeds don't is:
p^(nm-2) * (1-p)^2

Now the probability that these two are in two different rows is the same as 1 minus the probability that they are in the same row.
There are C(nm, 2) total possible two seed combiniations.
Of these, we can choose a row in n ways, and then we can choose 2 seeds in this row in C(m, 2) ways.
So the probability that any two seeds are in the same row is:
n*C(m, 2)/C(nm, 2)
= n*m!/[(m-2)!2!] / {(nm)!/[(nm-2)!2!]}
or
= n * m!(nm-2)!/[(m-2)!(nm)!]
= n * m*(m-1)/[(nm)*(nm-1)]
= (m-1)/(nm-1)

So the probability that the two seeds are not in the same row is
1 - (m-1)/(nm-1)
= (nm - m)/(nm - 1)
= m(n - 1)/(nm - 1)

So the total proability is
m * (n - 1) * p^(nm-2) * (1-p)^2 / (nm - 1)

{Note to Hal W
If you go back and evaluate your (unsimplified) answer:
{[p^m]^(n-2)} * [(p^[m-1]))*(1-p)]^2
it is just:
= p^(mn - 2m) * p^(2m - 2) * (1-p)^2
= p^(mn - 2m + 2m - 2) * (1-p)^2
= p^(mn - 2) * (1-p)^2
Which is the probability that any two seeds fail, and not two in different rows, so it can't be the final solution.}

Answer 2

The way i see it:

the probability that in a row of m seeds only 1 does not germinate is:

(p^(m-1)) x (1-p) x m

fully germinated x not fully germinated x configurations
there are m places where the ungerminated seed could go.

the probability that all seeds germinate in a row is:

p^m

therefore the prob that 2 rows contain 1 ungerminated seed is:

(p^(m(n-2))) x ((p^(m-1) x (1-p) x m)^2) x (n(n-1)/2)

fully germinated x not fully germinated x configurations
there are (n(n-1)/2) different configurations of the two rows that aren't fully germinated.

Answer 3

No, you don't need to take counting arrangements into account, other than to make sure that the two non-germinating seeds are in different rows.

Probability that all seeds in a row germinate is p^m
Probability of exactly one seed in a row failing to germinate is (p^[m-1]))*(1-p)

So the probability of (n-2) rows of fully germinated seeds and 2 rows with a single ungerminated seed is

{[p^m]^(n-2)} * [(p^[m-1]))*(1-p)]^2

which simplifies to

{p^(mn-2m)} * {p^(m-1) - p^m}^2

Answer 4

The answer I get is:
P[1 seed not germinating in row of m seeds ] = P[R] =
{ (m!) / [ (m - 1)! (1)! ] } * p^(m - 1) * (1 - p)
P[2 rows not germinating in n rows ] =
{ (n!) / [ (n - 2)! (2)! ] } * (p^m)^(n-2) * P[R]^2
=
n! * (p^m)^(n-2) * (m!)^2 * (p^(m-1))^2 * (1-p)^2 / [ ((m-1)!)^2 * (n-2)! * 2! * (1!)^2 ]

Answer 5

I don't know, but keep the packet and if too many fail check the stated germination rate and if it's an exaggeration take the packet back and ask for your money back

Answer 6

yeah with out a doubt, pass the spliff