probability of growing a healthy plant

**skorpiox** · August 2nd 2009, 06:25 PM

If a seed is planted, it has a 65% chance of growing into a healthy plant.

If 10 seeds are planted, what is the probability that exactly 2 don't grow?

I subtracted 1 from the probability that exaxtly 2 will grow , and my anwser is 1-.0042813781= 0.99572 ; is this correct , or do I have to do something else . Thanks in advance!!

**Chris L T521** · August 2nd 2009, 06:37 PM

Originally Posted by skorpiox

If a seed is planted, it has a 65% chance of growing into a healthy plant.

If 10 seeds are planted, what is the probability that exactly 2 don't grow?

I subtracted 1 from the probability that exaxtly 2 will grow , and my anwser is 0.99572; is this correct , or do I have to do something else . Thanks in advance!!

Note that if X represents the number of plants that grow, then it follows that $X$ has a $\mathcal{B}\left(10,.65\right)$ distribution.

If we want to know the probability of exactly two plants failing to grow, its the same as asking for the probability of exactly eight plants that succeeded to grow.

So it follows that the answer is $P\left(X=8\right)={10\choose 8}\left(.65\right)^{10}\left(.35\right)^{10-8}\approx0.0742$

Does this make sense?

**artvandalay11** · August 2nd 2009, 06:51 PM

This is a binomial distribution problem with n=10, p=.65

We want to know the probability that 8 plants grow, so k=8

$<br /> P(k)={{n}\choose{k}}p^k(1-p)^{n-k}<br />$

$<br /> P(8)={{10}\choose{8}}(.65)^8(1-.65)^{10-8}<br />$

${{n}\choose{k}}$ is notation for nCr in case you didnt know and you should be good from here

**skorpiox** · August 2nd 2009, 07:12 PM

thank you guys ; I got it. I guess I was thinking about normal distribution

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