A shopkeeper sells three types of flower seeds: $A_{1}$ , $A_{2}$ , and $A_{3}$ . They are sold as a mixture where the proportions are 4:4:2, respectively. The germination rates of the three types of seeds are 45%, 60%, and 35%. Calculate the probability:

(i) Of a randomly chosen seed to germinate

(ii) That it will not germinate given that the seed is of type $A_{3}$

(iii) That it is of type $A_{2}$ given that a randomly chosen seed does not germinate.

4 Views | Posted 3 months ago

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Answered by

alok kumar singh | Contributor-Level 10

3 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t A_{1} : A_{2} : A_{3} = 4 : 4 : 2 \\ ∴ P (A_{1}) = \frac{4}{10}, P (A_{2}) = \frac{4}{10} a n d P (A_{3}) = \frac{2}{10} w h e r e A_{1}, A_{2} a n d A_{3} a r e t h e t h r e e t y p e s o f s e e d s . \\ L e t E b e t h e e v e n t s t h a t a s e e d a n d \bar{E} b e t h e e v e n t s t h a t a s e e d d o e s n o t \\ ∴ P (\frac{E}{A_{1}}) = \frac{45}{100}, P (\frac{E}{A_{2}}) = \frac{60}{100} a n d P (\frac{E}{A_{3}}) = \frac{35}{100} \\ a n d P (\frac{\bar{E}}{A_{1}}) = \frac{55}{100}, P (\frac{\bar{E}}{A_{2}}) = \frac{40}{100} a n d P (\frac{\bar{E}}{A_{3}}) = \frac{65}{100} \\ (i) P (E) = P (A_{1}) . P (\frac{E}{A_{1}}) + P (A_{2}) . P (\frac{E}{A_{2}}) + P (A_{3}) . P (\frac{E}{A_{3}}) \\ = \frac{4}{10} . \frac{45}{100} + \frac{4}{10} . \frac{60}{100} + \frac{2}{10} . \frac{35}{100} \\ = \frac{180}{1000} + \frac{240}{1000} + \frac{70}{1000} = \frac{490}{1000} = 0.49 \\ (i i) P (\bar{E} / A_{3}) = 1 - P (E / A_{3}) = 1 - \frac{35}{100} = \frac{65}{100} = 0.65 \\ (i i i), B a y e s' T h e o r e m, w e g e t \\ P (A_{2} / \bar{E}) = \frac{P (A_{2}) . P (\bar{E} / A_{2})}{P (A_{1}) . P (\bar{E} / A_{1}) + P (A_{2}) . P (\bar{E} / A_{2}) + P (A_{3}) . P (\bar{E} / A_{3})} \\ = \frac{\frac{4}{10} . \frac{40}{100}}{\frac{4}{10} . \frac{55}{100} + \frac{4}{10} . \frac{40}{100} + \frac{2}{10} . \frac{65}{100}} = \frac{\frac{160}{1000}}{\frac{220}{1000} + \frac{160}{1000} + \frac{130}{1000}} \\ = \frac{160}{220 + 160 + 130} = \frac{160}{510} = \frac{16}{51} = 0.314 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{16}{51} o r 0.314 \end{array}$

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