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Question Number 194490 by pete last updated on 08/Jul/23

$$\mathrm{A}\mathrm{research}\mathrm{station}\mathrm{supplies}\mathrm{three}\mathrm{varieties}$$$$\mathrm{of}\mathrm{seeds}\mathrm{S}1,\mathrm{S}2\mathrm{and}\mathrm{S}3\mathrm{in}\mathrm{the}\mathrm{ratio}4:2:1.$$$$\mathrm{The}\mathrm{probabilities}\mathrm{of}\mathrm{germination}\mathrm{of}$$$$\mathrm{S}1,\mathrm{S}2\mathrm{and}\mathrm{S}3\mathrm{are}50\mathrm{\%},60\mathrm{\%}\mathrm{and}80\mathrm{\%}$$$$\mathrm{respectively}.\mathrm{Find}\mathrm{the}\mathrm{probability}\mathrm{that}$$$$\mathrm{a}\mathrm{seed}\mathrm{selected}\mathrm{at}\mathrm{random}\mathrm{will}\mathrm{germinate}.$$

Answered by MM42 last updated on 08/Jul/23

$$$$$$p\left(g\right)=p\left(s_1\right)\times p(g\mid s_1)+p\left(s_2\right)\times p(g\mid s_2)+p\left(s_3\right)\times p(g\mid s_3)$$$$=\frac{4}{7}\times \frac{1}{2}+\frac{2}{7}\times \frac{3}{5}+\frac{1}{7}\times \frac{4}{5}$$$$=\frac{4}{7}$$$$$$

Commented by pete last updated on 09/Jul/23

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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