binomial distribution
we want chance of exactly 5 of 6 making it
p(grow) = .7
so p(not) = .3
P(6,5) =C(6,5) .7^6 * .3^1
= 6 (.7)^6 * .3
= .212
if a seed is planted, it has a 70% chance of growing into a healthy plant.if 6 seeds are planted, what is the probability that exactly 1 doesn't grow?
2 answers
Let event
G=seed grows into a healthy plant (success)
~G=seed does not grow (failure)
We use the binomial distribution where
N=6,
p=0.70 (probability of success)
n=5 (5 successes)
C(N,n)=N!/((N-n)!n!)
then by binomial theorem
P(n=5)=C(6,5)p^5(1-p)^1
=6!/(1!5!)0.7^5(0.3^1)
=.3025
G=seed grows into a healthy plant (success)
~G=seed does not grow (failure)
We use the binomial distribution where
N=6,
p=0.70 (probability of success)
n=5 (5 successes)
C(N,n)=N!/((N-n)!n!)
then by binomial theorem
P(n=5)=C(6,5)p^5(1-p)^1
=6!/(1!5!)0.7^5(0.3^1)
=.3025