If a seed is planted, it has a 75% chance of growing into a healthy plant.

If 6 seeds are planted, what is the probability that exactly 1 doesn't grow?

9.72405%

Step-by-step explanation:

Binomial Probability

(N choose k) p^k (1-p)^(n-k)

N=7  seeds planted

p= 100% - 70% = 30% = 0.3 <--- we are interested in the plant NOT growing

(1-p) = 70% = 0.7  <--- 70% chance the plant will survive and grow

k=4  <--- we want four of them to fail

The probability is:

 (7 choose 4) * (0.3)^4  (0.7)^3 =

 7!/(4!3!) (0.3)^4 (0.7)^3  =

 (7*6*5/3*2) (0.3)^4 (0.7)^3 =

 7*5 (0.3)^4 (0.7)^3 =

 35 * 0.0081 * 0.343 = 0.0972405 = 9.72405%

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )