If a seed is planted, it has a 90% chance of growing into a healthy plant.

If 9 seeds are planted, what is the probability that exactly 2 don't grow?

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- Jul 2nd 2014, 09:57 AMroy035117Probability
If a seed is planted, it has a 90% chance of growing into a healthy plant.

If 9 seeds are planted, what is the probability that exactly 2 don't grow?

- Jul 2nd 2014, 01:06 PMebainesRe: Probability
For exactly 2 plants to not grow, that means 7 do grow. The probability of an individual plant growing is p=0.9. Conversely the probability of an individual plant not growing is q=1-p = 0.1. The probability of exactly k successes in n trials, where the probability of the success of any one trial is p and the probability of failure on any one trial is q=1-p, is given by the Binomial Probability Formula:

$\displaystyle p(k \ successes \ in\ n \ trials) = \dbinom{n}{k} p^k q^{n-k}$

Here p = 0.9, q=0.1, k = 7, and n=9.

Can you take it from here? What do you get for a final answer? - Jul 2nd 2014, 01:18 PMHallsofIvyRe: Probability
The probability that the

**first**doesn't grow is 0.10. The probability that the**second**doesn't grow is also 0.10. The probability each of the next 0.90. The probability that the first two don't grow and the next 7 do not grow is (0.10)(0.10)(0.90)(0.90)(0.90)(0.90)(0.90)(0.90)(0 .90)= $\displaystyle (0.10)^2(0.90)^7$. The same kind of argument will show that two not growing and 7 growing in**any**order is also $\displaystyle (0.10)^2(0.90)^7$. But then we can show that there are $\displaystyle \frac{9!}{7! 2!}$ such different orders so the probability of "seven grow and two don't" is $\displaystyle \frac{9!}{7! 2!}(0.10)^2(0.90)^7$.