Conditional Probability (kinda) Question?

Question

How do you solve a problem of this type? The best way I can explain is with an example:

A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probabilty of .08 of being dead, a probability of .19 of exhibiting slow growth, a probability of .42 of exhibiting medium growth, and a probability of .31 of exhibiting strong growth.

Calculate the probability that after one month:
a)Exactly three seedlings are dead, exactly four exhibit slow growth and exactly six exhibit medium growth.
b)Exactly five seedlings are dead, exactly five exhibit slow growth, and exactly seven exhibit strong growth.
c)No more than two seedlings are dead.


Please don't just give me the answer (this isn't the problem on my assignment so that won't do me any good), I need an explanation on how to solve - thanks in advance!

Smiley, I don't think you're right, but you pointed me in the right direction: I basically took the (x choose k) * p^k * (1-p)^(x-k) equation, did that for all three individual successes and multiplied them - my answer's .00233.

And for the second, .0006807.

Answer 1

I think that you can just multiply the individual probabilities together. Just remeber that you have to account for all 22 plants.

So example a)
3 dead 4 slow 6 medium (9 have to have strong growth)
(.08)^3 (.19)^4 (.42)^6 (.31)^9

b)
5 dead 5 slow 7 strong (5 have medium growth)
(.08)^5 (.19)^5 (.31)^7 (.42)^5

c) Assume that exactly 2 are dead. Then you have 20 other plants that could be anything but dead so the probability is .92 for the others.
(.08)^2 (.92)^20

I may be thinking too simplistically, but that's what I would try.

Good Luck. Hope this makes sense and helps.

Answer 2

I forget how to do a and b.

c)
Prob that none die + that only one dies + that exactly 2 dies.