No, this is an example of Binomial Distribution.
You are planting 7 plants, and you want the prob that exactly 1 does not grow.
Suppose it is the 3rd plant which did not grow.
GGNGGGG, where G stands for grow, N for not grow.
That specific probability is
(.7)(.7)(.3)(.7)(.7)(.7)(.7)
= (.7^6)(.3)
but it could have been 7 different arrangements of G's and N
e.g.
GGGGGNG is another case
So you the prob as
7 (.7)^(.3)
= appr .247
or
C(7,1) (.7)^6 (.3) = appr .247
If a seed is planted, it has a 70% chance of growing into a healthy plant.
If 7 seeds are planted, what is the probability that exactly 1 doesn't grow?
Here What I did I thought its premutation but some reason I got that wrong, I multiple by (0.10) and (0.70) of the seeds I still got it wrong. Where did I got it wrong
1 answer
1 answer