Question

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.

• A. a
• B. b
• C. c
• D. d

Answer 1

The Correct Option is A

We have given, $A_{1} : A_{2} : A_{3} =4 : 4 : 2$ $\therefore P\left(A_{1}\right)=\frac{4}{10}, P\left(A_{2}\right) =\frac{4}{10}$ 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. Then $P\left(E| A_{1}\right)=\frac{45}{100}, P\left(E |A_{2}\right)=\frac{60}{100}$ and $P\left(E |A_{3}\right)=\frac{35}{100}$ and $P\left(\bar{E} |A_{1}\right)=1-P\left(E| A_{1}\right)=\frac{55}{100}$, $P\left(\bar{E} |A_{2}\right)=1-p\left(E| A_{2}\right)=\frac{40}{100}$ and $P\left(\bar{E} |A_{3}\right)=1-P\left(E| A_{3}\right)=\frac{65}{100}$ $\left(i\right)$ $P\left(E\right)=P\left(A_{1}\right). P\left(E| A_{1}\right)$ $+P\left(A_{2}\right).P\left(E |A_{2}\right)+P\left(A_{3}\right). P\left(E |A_{3}\right)$ $=\frac{4}{10}\cdot\frac{45}{100}+\frac{4}{10}\cdot\frac{60}{100}+\frac{2}{10}\cdot\frac{35}{100}$ [Substituting above values] $=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49$ $\left(ii\right)$ $P\left(\bar{E}| A_{3}\right)=1-P\left(E |A_{3}\right)=1-\frac{35}{100}=\frac{65}{100}=0.65$ $\left(iii\right) P\left(A_{2} |\bar{E}\right)=\frac{P\left(A_{2}\right). P\left(\bar{E} |A_{2}\right)}{P\left(A_{1}\right)\cdot P\left(\bar{E} |A_{1}\right)+P\left(A_{2}\right)\cdot P\left(\bar{E}| A_{2}\right)+P\left(A_{3}\right)\cdot P\left(\bar{E} |A_{3}\right)}$ $=\frac{\frac{4}{10}\cdot\frac{40}{100}}{\frac{4}{10}\cdot\frac{55}{100}+\frac{4}{10}\cdot\frac{40}{100}+\frac{2}{10}\cdot\frac{65}{100}}$ $=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}}$ $=\frac{160 /1000}{510 / 1000}=\frac{16}{51}$ $=0.314$