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%

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation:

the answer is 13 cm