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Question

Mathematics Question on Conditional Probability

A shopkeeper sells three types of flower seeds A1A_{1}, A2A_{2} and A3A_{3}. They are sold as a mixture, where the proportions are 4:4:24:4:2, respectively. The germination rates of the three types of seeds are 45%45\%, 60%60\% and 35%35\%. calculate the probability (i)(i) of a randomly chosen seed to germinate. (ii)(ii) that it will not germinate given that the seed is of type A3A_{3}. (iii)( iii) that it is of the type A2A_{2} given that a randomly chosen seed does not germinate.

A

a

B

b

C

c

D

d

Answer

a

Explanation

Solution

We have given, A1:A2:A3=4:4:2A_{1} : A_{2} : A_{3} =4 : 4 : 2 P(A1)=410,P(A2)=410\therefore P\left(A_{1}\right)=\frac{4}{10}, P\left(A_{2}\right) =\frac{4}{10} and P(A3)=210P\left(A_{3}\right)=\frac{2}{10} where A1A_{1}, A2A_{2} and A3A_{3} denote the three types of flower seeds. Let EE be the event that a seed germinates. Then P(EA1)=45100,P(EA2)=60100P\left(E| A_{1}\right)=\frac{45}{100}, P\left(E |A_{2}\right)=\frac{60}{100} and P(EA3)=35100P\left(E |A_{3}\right)=\frac{35}{100} and P(EˉA1)=1P(EA1)=55100P\left(\bar{E} |A_{1}\right)=1-P\left(E| A_{1}\right)=\frac{55}{100}, P(EˉA2)=1p(EA2)=40100P\left(\bar{E} |A_{2}\right)=1-p\left(E| A_{2}\right)=\frac{40}{100} and P(EˉA3)=1P(EA3)=65100P\left(\bar{E} |A_{3}\right)=1-P\left(E| A_{3}\right)=\frac{65}{100} (i)\left(i\right) P(E)=P(A1).P(EA1)P\left(E\right)=P\left(A_{1}\right). P\left(E| A_{1}\right) +P(A2).P(EA2)+P(A3).P(EA3)+P\left(A_{2}\right).P\left(E |A_{2}\right)+P\left(A_{3}\right). P\left(E |A_{3}\right) =41045100+41060100+21035100=\frac{4}{10}\cdot\frac{45}{100}+\frac{4}{10}\cdot\frac{60}{100}+\frac{2}{10}\cdot\frac{35}{100} [Substituting above values] =1801000+2401000+701000=4901000=0.49=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49 (ii)\left(ii\right) P(EˉA3)=1P(EA3)=135100=65100=0.65P\left(\bar{E}| A_{3}\right)=1-P\left(E |A_{3}\right)=1-\frac{35}{100}=\frac{65}{100}=0.65 (iii)P(A2Eˉ)=P(A2).P(EˉA2)P(A1)P(EˉA1)+P(A2)P(EˉA2)+P(A3)P(EˉA3)\left(iii\right) P\left(A_{2} |\bar{E}\right)=\frac{P\left(A_{2}\right). 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)} =4104010041055100+41040100+21065100=\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}} =16010002201000+1601000+1301000=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}} =160/1000510/1000=1651=\frac{160 /1000}{510 / 1000}=\frac{16}{51} =0.314=0.314