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?
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?
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 $\displaystyle \frac{9!}{7! 2!}(0.10)^2(0.90)^7$.