Question

A shopkeeper sells three types of flower seeds A1, A2 and A3. 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% respectively .

(i) Find the probability of a randomly chosen seed to germinate

(ii) If a randomly chosen seed does not germinate, find the probability that it is of the type A2. ​

Answer 1

Given :

We have, $\[{A_1}:{A_2}:{A_3} = 4:4:2\]$

where, A₁, A₂, A₃ denote the three types of flower seeds.

$\[P\left( {{E \mathord{\left/ {\vphantom {E {{A_1}}}} \right. \kern-\nulldelimiterspace} {{A_1}}}} \right) = \frac{{45}}{{100}}\]$, $\[P\left( {{E \mathord{\left/ {\vphantom {E {{A_2}}}} \right. \kern-\nulldelimiterspace} {{A_2}}}} \right) = \frac{{60}}{{100}}\]$, $\[P\left( {{E \mathord{\left/ {\vphantom {E {{A_1}}}} \right. \kern-\nulldelimiterspace} {{A_1}}}} \right) = \frac{{35}}{{100}}\]$                 and

$\[P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_1}}}} \right. \kern-\nulldelimiterspace} {{A_1}}}} \right) = \frac{{55}}{{100}}\]$,  $\[P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_2}}}} \right. \kern-\nulldelimiterspace} {{A_2}}}} \right) = \frac{{40}}{{100}}\]$, $\[P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_3}}}} \right. \kern-\nulldelimiterspace} {{A_3}}}} \right) = \frac{{65}}{{100}}\]$

where, E be the event that seed germinate and E' be the event that the seed doesn't germinate.

To Find :

We have to find the probability,

1. of a randomly chosen seed to germinate
2. that it is of type A₂ given that the randomly chosen seed doesn't germinate.

Solution:

1. $\[P\left( E \right) = P\left( {{A_1}} \right) \cdot P\left( {{E \mathord{\left/ {\vphantom {E {{A_1}}}} \right. \kern-\nulldelimiterspace} {{A_1}}}} \right) + P\left( {{A_2}} \right) \cdot P\left( {{E \mathord{\left/ {\vphantom {E {{A_2}}}} \right. \kern-\nulldelimiterspace} {{A_2}}}} \right) + P\left( {{A_3}} \right) \cdot P\left( {{E \mathord{\left/ {\vphantom {E {{A_3}}}} \right. \kern-\nulldelimiterspace} {{A_3}}}} \right)\]$

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

                   $\[ = \frac{{180}}{{1000}} + \frac{{240}}{{1000}} + \frac{{70}}{{1000}}\]$

                   $\[ = \frac{{490}}{{1000}}\]$

                   $\[ = 0 \cdot 49\]$

Hence, the probability of a randomly chosen seed to germinate is 0·49.

    2.  $\[P\left( {{{{A_2}} \mathord{\left/ {\vphantom {{{A_2}} {{E^'}}}} \right. \kern-\nulldelimiterspace} {{E^'}}}} \right) = \frac{{No.\,of\,favourable\,outcomes}}{{Total\,number\,of\,outcomes}}\]$

          $\[\begin{array}{c}No.\,of\,favourable\,outcomes = P\left( {{A_2}} \right) \cdot P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_2}}}} \right. \kern-\nulldelimiterspace} {{A_2}}}} \right)\\ = \frac{4}{{10}} \cdot \frac{{40}}{{100}}\\\\ = \frac{4}{{25}}\end{array}\]$

    $\[Total\,number\,of\,outcomes = P\left( {{A_1}} \right) \cdot P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_1}}}} \right. \kern-\nulldelimiterspace} {{A_1}}}} \right) + P\left( {{A_2}} \right) \cdot P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_2}}}} \right. \kern-\nulldelimiterspace} {{A_2}}}} \right) + P\left( {{A_3}} \right) \cdot P\left( {{{{E^'}} \mathord{\left/ {\vphantom {{{E^'}} {{A_3}}}} \right. \kern-\nulldelimiterspace} {{A_3}}}} \right)\]$        

                                            $\[\begin{array}{c} = \frac{4}{{10}} \cdot \frac{{55}}{{100}} + \frac{4}{{10}} \cdot \frac{{40}}{{100}} + \frac{2}{{10}} \cdot \frac{{65}}{{100}}\\ \\ = \frac{{51}}{{100}}\end{array}\]$

                             $\[\therefore P\left( {{{{A_2}} \mathord{\left/ {\vphantom {{{A_2}} {{E^'}}}} \right. \kern-\nulldelimiterspace} {{E^'}}}} \right) = \frac{{16}}{{51}}\]$

Hence, the probability that it is of type A₂ given that the randomly chosen seed doesn't germinate is  $\frac{16}{51}$