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

If 6 seeds are planted, what is the probability that exactly 2 don't grow?

Answer:

[tex]\displaystyle\frac{19,683}{200,000}\text{ or }\approx 9.84\%[/tex]

Step-by-step explanation:

For each planted seed, there is a 90% chance that it grows into a healthy plant, which means that there is a [tex]100\%-90\%=10\%[/tex] chance it does not grow into a healthy plant.

Since we are planting 6 seeds, we want to choose 2 that do not grow and 4 that do grow:

[tex]\displaystyle \frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}[/tex]

However, this is only one case that meets the conditions. We can choose any 2 out of the 6 seeds to be the ones that don't grow into a healthy plant, not just the first and second ones. Therefore, we need to multiply this by number of ways we can choose 2 things from 6 (6 choose 2):

[tex]\displaystyle \binom{6}{2}=\frac{6\cdot 5}{2!}=\frac{30}{2}=15[/tex]

Therefore, we have:

[tex]\displaystyle\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{\frac{19,683}{200,000}}\approx 9.84\%[/tex]

Answer:

[tex] {?}^{?} However, this is only one case that meets the conditions. We can choose any 2 out of the 6 seeds to be the ones that don't grow into a healthy plant, not just the first and second ones. Therefore, we need to multiply this by number of ways we can choose 2 things from 6 (6 choose 2):

\displaystyle \binom{6}{2}=\frac{6\cdot 5}{2!}=\frac{30}{2}=15(26)=2!6⋅5=230=15

Therefore, we have:

\begin{gathered}\displaystyle\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{\frac{19,683}{200,000}}\approx 9.84\%\end{gathered}P(exactly 2 don’t grow)=101⋅101⋅109⋅109⋅109⋅109⋅(26),P(exactly 2 don’t grow)=101⋅101⋅109⋅109⋅109⋅109⋅15,P(exactly 2 don’t grow)=200,00019,683≈9.84%

[/tex]

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