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.

	(i)	(ii)	(iii)
(a)	$0.49\,\,\,$	$0.65\,\,\,$	$0.314\,\,\,$
(b)	$0.49$	$0.75$	$0.314$
(c)	$0.65$	$0.49$	$0.314$
(d)	$0.49$	$0.314$	$0.65$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (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]
$=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49$
$\left(ii\right)$ $P\left(\bar{E}| A_{3}\right)=1-P\left(E |A_{3}\right)=1-\frac{35}{100}=\frac{65}{100}=0.65$
$\left(iii\right) P\left(A_{2} |\bar{E}\right)=\frac{P\left(A_{2}\right). P\left(\bar{E} |A_{2}\right)}{P\left(A_{1}\right)\cdot P\left(\bar{E} |A_{1}\right)+P\left(A_{2}\right)\cdot P\left(\bar{E}| A_{2}\right)+P\left(A_{3}\right)\cdot P\left(\bar{E} |A_{3}\right)}$
$=\frac{\frac{4}{10}\cdot\frac{40}{100}}{\frac{4}{10}\cdot\frac{55}{100}+\frac{4}{10}\cdot\frac{40}{100}+\frac{2}{10}\cdot\frac{65}{100}}$
$=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}}$
$=\frac{160 /1000}{510 / 1000}=\frac{16}{51}$
$=0.314$