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Case study (4 Marks)

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Question 14 Marks
Read the text carefully and answer the questions:
A shopkeeper sells three types of flower seeds $\mathrm{A}_{1}, \mathrm{~A}_{2}$, and $\mathrm{A}_{3}$. 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 $35 \%$ respectively.
(a) Calculate the probability of randomly chosen seed to germinate.
(b) Calculate the probability that it is of the type $\mathrm{A}_{2}$ given that randomly chosen seed does not germinate.
(c) Calculate the probability that it will not germinate given that the seed is of type $\mathrm{A}_{1}$.
Answer
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 $35 \%$ respectively.
(i) Here, $\mathrm{P}\left(\mathrm{A}_{1}\right)=\frac{4}{10}, \mathrm{P}\left(\mathrm{A}_{2}\right)=\frac{4}{10}, \mathrm{P}\left(\mathrm{A}_{3}\right)=\frac{2}{10}$,
and $\mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{1}\right)=\frac{45}{100}, \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{2}\right)=\frac{60}{100}, \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{3}\right)=\frac{35}{100}$
where $G$ is the event that seeds germinate.
$\mathrm{P}(\mathrm{G})=\mathrm{P}\left(\mathrm{A}_{1}\right) \cdot \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{1}\right)+\mathrm{P}\left(\mathrm{A}_{2}\right) \cdot \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{2}\right)+\mathrm{P}\left(\mathrm{A}_{3}\right) \cdot \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{3}\right)$
$=\frac{4}{10} \times \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \times \frac{35}{100}=\frac{490}{1000}=0.49$
(ii) Here, $\mathrm{P}\left(\mathrm{A}_{1}\right)=\frac{4}{10}, \mathrm{P}\left(\mathrm{A}_{2}\right)=\frac{4}{10}, \mathrm{P}\left(\mathrm{A}_{3}\right)=\frac{2}{10}$,
and $\mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{1}\right)=\frac{45}{100}, \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{2}\right)=\frac{60}{100}, \mathrm{P}\left(\mathrm{G} \mid \mathrm{A}_{3}\right)=\frac{35}{100}$
where $G$ is the event that seeds germinate.
Required probability $=P\left(\mathrm{~A}_{2} \mid \mathrm{G}^{\prime}\right)=\frac{\mathrm{P}\left(\mathrm{A}_{2}\right) \cdot \mathrm{P}\left(\mathrm{G}^{\prime} \mid \mathrm{A}_{2}\right)}{\mathrm{P}\left(\mathrm{A}_{1}\right) \cdot \mathrm{P}\left(\mathrm{G}^{\prime} \mid \mathrm{A}_{1}\right)+\mathrm{P}\left(\mathrm{A}_{2}\right) \cdot \mathrm{P}\left(\mathrm{G}^{\prime} \mid \mathrm{A}_{2}\right)+\mathrm{P}\left(\mathrm{A}_{3}\right) \cdot \mathrm{P}\left(\mathrm{G}^{\prime} \mid \mathrm{A}_{3}\right)}$
$
=\frac{\frac{4}{10} \times \frac{40}{100}}{\frac{4}{10} \times \frac{55}{100}+\frac{4}{10} \times \frac{40}{100}+\frac{2}{10} \times \frac{65}{100}}=\frac{160}{510}=\frac{16}{51}=0.314
$
(iii) P (seeds of type $\mathrm{A}_{1}$ will not germinate) $=1-\frac{45}{100}=\frac{55}{100}$.
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Question 24 Marks
Read the text carefully and answer the questions:
Two students Ankit and Vinod appeared in an examination. The probability that Ankit will qualify the examination is 0.05 and that Vinod will qualify is 0.10 . The probability that both will qualify is 0.02 .
(a) Find the probability that atleast one of them will qualify the exam.
(b) Find the probability that atleast one of them will not qualify the exam.
(c) Find the probability that both Ankit and Vinod will not qualify the exam.
Answer
Two students Ankit and Vinod appeared in an examination. The probability that Ankit will qualify the examination is 0.05 and that Vinod will qualify is 0.10 . The probability that both will qualify is 0.02 .
(i) Let $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ denotes the events that Ankit and Vinod will respectively qualify the exam.
$
\begin{aligned}
& P\left(E_{1} \cup E_{2}\right)=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)-\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right) \\
& =0.05+0.10-0.02=0.13
\end{aligned}
$
(ii) Let $E_{1}$ and $E_{2}$ denotes the events that Ankit and Vinod will respectively qualify the exam.
Probability of atleast one of them does not qualify
$=P\left(E_{1}^{\prime} \cup E_{2}^{\prime}\right)=P\left(\left(E_{1} \cap E_{2}\right)^{\prime}\right)$
$=1-\mathrm{p}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)=1-0.02=0.98$
(iii)Let $E_{1}$ and $E_{2}$ denotes the events that Ankit and Vinod will respectively qualify the exam.
$
\begin{aligned}
& =P\left(E_{1}^{\prime} \cap E_{2}^{\prime}\right)=P\left(\left(E_{1} \cup E_{2}\right)^{\prime}\right) \\
& =1-\mathrm{P}\left(\mathrm{E}_{1} \cup \mathrm{E}_{2}\right)=1-0.13=0.87
\end{aligned}
$
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Question 34 Marks
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Question 44 Marks
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