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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?

User Sudonym
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2 Answers

5 votes

Answer:


{?}^(?) However, this is only one case that meets the conditions. We can choose&nbsp;any&nbsp;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):</p><p></p><p>\displaystyle \binom{6}{2}=(6\cdot 5)/(2!)=(30)/(2)=15(26)=2!6⋅5=230=15</p><p></p><p>Therefore, we have:</p><p></p><p>\begin{gathered}\displaystyle\\P(\text{exactly 2 don't grow})=(1)/(10)\cdot (1)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=(1)/(10)\cdot (1)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{(19,683)/(200,000)}\approx 9.84\%\end{gathered}P(exactly&nbsp;2&nbsp;don’t&nbsp;grow)=101⋅101⋅109⋅109⋅109⋅109⋅(26),P(exactly&nbsp;2&nbsp;don’t&nbsp;grow)=101⋅101⋅109⋅109⋅109⋅109⋅15,P(exactly&nbsp;2&nbsp;don’t&nbsp;grow)=200,00019,683≈9.84%</p><p></p><p>

User Yellow Days
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4.2k points
2 votes

Answer:


\displaystyle(19,683)/(200,000)\text{ or }\approx 9.84\%

Explanation:

For each planted seed, there is a 90% chance that it grows into a healthy plant, which means that there is a
100\%-90\%=10\% 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:


\displaystyle (1)/(10)\cdot (1)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot (9)/(10)\cdot (9)/(10)

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}=(6\cdot 5)/(2!)=(30)/(2)=15

Therefore, we have:


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

User Ebram
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4.3k points