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Question Number 225568 by necx122 last updated on 03/Nov/25
A farmer produces seeds in packets for sale. The probability that a seed selected at random will grow is 0.8. If there are 20 seeds, what is the probability that less than 2 will not grow?
$${A}\:{farmer}\:{produces}\:{seeds}\:{in}\:{packets} \\ $$$${for}\:{sale}.\:{The}\:{probability}\:{that}\:{a}\:{seed} \\ $$$${selected}\:{at}\:{random}\:{will}\:{grow}\:{is}\:\mathrm{0}.\mathrm{8}. \\ $$$${If}\:{there}\:{are}\:\mathrm{20}\:{seeds},\:{what}\:{is}\:{the} \\ $$$${probability}\:{that}\:{less}\:{than}\:\mathrm{2}\:{will}\:{not} \\ $$$${grow}? \\ $$
Commented by necx122 last updated on 03/Nov/25
Please help
$${Please}\:{help} \\ $$
Answered by mr W last updated on 04/Nov/25
the probability that a seed grows is 0.8. that means the probability that it doesn′t grow is 0.2. this is a so−called binomial distribution problem. the probability that r from 20 seeds don′t grow is p_r = (((20)),(r) )×0.2^r ×0.8^(20−r) “less than 2 seeds” means zero seed or one seed, i.e. r=0 or 1. p_0 = (((20)),(0) )×0.2^0 ×0.8^(20) p_1 = (((20)),(1) )×0.2^1 ×0.8^(19) totally p_0 +p_1 = (((20)),(0) )×0.2^0 ×0.8^(20) + (((20)),(1) )×0.2^1 ×0.8^(19) ≈0.069
$${the}\:{probability}\:{that}\:{a}\:{seed}\:{grows}\:{is} \\ $$$$\mathrm{0}.\mathrm{8}.\:{that}\:{means}\:{the}\:{probability}\:{that}\: \\ $$$${it}\:{doesn}'{t}\:{grow}\:{is}\:\mathrm{0}.\mathrm{2}. \\ $$$${this}\:{is}\:{a}\:{so}−{called}\:{binomial} \\ $$$${distribution}\:{problem}.\:{the}\:{probability} \\ $$$${that}\:{r}\:{from}\:\mathrm{20}\:{seeds}\:{don}'{t}\:{grow}\:{is} \\ $$$${p}_{{r}} =\begin{pmatrix}{\mathrm{20}}\\{{r}}\end{pmatrix}×\mathrm{0}.\mathrm{2}^{{r}} ×\mathrm{0}.\mathrm{8}^{\mathrm{20}−{r}} \\ $$$$“{less}\:{than}\:\mathrm{2}\:{seeds}''\:{means}\:{zero}\:{seed} \\ $$$${or}\:{one}\:{seed},\:{i}.{e}.\:{r}=\mathrm{0}\:{or}\:\mathrm{1}. \\ $$$${p}_{\mathrm{0}} =\begin{pmatrix}{\mathrm{20}}\\{\mathrm{0}}\end{pmatrix}×\mathrm{0}.\mathrm{2}^{\mathrm{0}} ×\mathrm{0}.\mathrm{8}^{\mathrm{20}} \\ $$$${p}_{\mathrm{1}} =\begin{pmatrix}{\mathrm{20}}\\{\mathrm{1}}\end{pmatrix}×\mathrm{0}.\mathrm{2}^{\mathrm{1}} ×\mathrm{0}.\mathrm{8}^{\mathrm{19}} \\ $$$${totally}\: \\ $$$${p}_{\mathrm{0}} +{p}_{\mathrm{1}} = \\ $$$$\begin{pmatrix}{\mathrm{20}}\\{\mathrm{0}}\end{pmatrix}×\mathrm{0}.\mathrm{2}^{\mathrm{0}} ×\mathrm{0}.\mathrm{8}^{\mathrm{20}} +\begin{pmatrix}{\mathrm{20}}\\{\mathrm{1}}\end{pmatrix}×\mathrm{0}.\mathrm{2}^{\mathrm{1}} ×\mathrm{0}.\mathrm{8}^{\mathrm{19}} \\ $$$$\approx\mathrm{0}.\mathrm{069} \\ $$
Commented by necx122 last updated on 03/Nov/25
This is really understandable. Thank you, sir.
$${This}\:{is}\:{really}\:{understandable}.\:{Thank} \\ $$$${you},\:{sir}. \\ $$
Commented by mr W last updated on 03/Nov/25
with pleasure!
$${with}\:{pleasure}! \\ $$

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