class 12 maths probability

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 the type ${A_2}$ given that a randomly chosen seed does not germinate.

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📘 Probability NCERT,Exemp,Q.43,Page.276 LA

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 the type ${A_2}$ given that a randomly chosen seed does not germinate.

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

We have, ${A_1}:{A_2}:{A_3} = 4:4:2$
$P\left( {{A_1}} \right) = \frac{4}{{10}},P\left( {{A_2}} \right) = \frac{4}{{10}}$

and $P\left( {{A_3}} \right) = \frac{2}{{10}}$

where ${A_1},{A_2}$ and ${A_3}$ denote the three types of flower seeds.

Let $E$ be the event that a seed germinates and $\bar E$

be the event that a seed does not germinate. $\therefore$ $\therefore$ $\therefore$ $P\left( {E/{A_1}} \right) = \frac{{45}}{{100}},P\left( {E/{A_2}} \right) = \frac{{60}}{{100}}$

and $P\left( {E/{A_3}} \right) = \frac{{35}}{{100}}$

and $P\left( {\bar E/{A_1}} \right) = \frac{{55}}{{100}},P\left( {\bar E/{A_2}} \right) = \frac{{40}}{{100}}$ and $P\left( {\bar E/{A_3}} \right) = \frac{{65}}{{100}}$

(i) $\therefore$ $\quad P(E) = P\left( {{A_1}} \right) \cdot P\left( {E/{A_1}} \right) + P\left( {{A_2}} \right) \cdot P\left( {E/{A_2}} \right) + P\left( {{A_3}} \right) \cdot P\left( {E/{A_3}} \right)$

$= \frac{4}{{10}} \cdot \frac{{45}}{{100}} + \frac{4}{{10}} \cdot \frac{{60}}{{100}} + \frac{2}{{10}} \cdot \frac{{35}}{{100}}$

$= \frac{{180}}{{1000}} + \frac{{240}}{{1000}} + \frac{{70}}{{1000}} = \frac{{490}}{{1000}} = 0.49$

(ii) $P\left( {\bar E/{A_3}} \right) = 1 - P\left( {E/{A_3}} \right) = 1 - \frac{{35}}{{100}} = \frac{{65}}{{100}}$ [as given above]

(iii) $P\left( {{A_2}/\bar E} \right) = \frac{{P\left( {{A_2}} \right) \cdot P\left( {\bar E/{A_2}} \right)}}{{P\left( {{A_1}} \right) \cdot P\left( {\bar E/{A_1}} \right) + P\left( {{A_2}} \right) \cdot P\left( {\bar E/{A_2}} \right) + P\left( {{A_3}} \right) \cdot P\left( {\bar E/{A_3}} \right)}}$

$= \frac{{\frac{4}{{10}} \cdot \frac{{40}}{{100}}}}{{\frac{4}{{10}} \cdot \frac{{55}}{{100}} + \frac{4}{{10}} \cdot \frac{{40}}{{100}} + \frac{2}{{10}} \cdot \frac{{65}}{{100}}}} = \frac{{\frac{{160}}{{1000}}}}{{\frac{{220}}{{1000}} + \frac{{160}}{{1000}} + \frac{{130}}{{1000}}}}$

$= \frac{{160/1000}}{{510/1000}} = \frac{{16}}{{51}} = 0.313725 = 0.314$

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