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  • 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%. Calculate the probability (i) of a randomly cho

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%. Calculate the probability
(i) of a randomly chosen seed to germinate
(ii) that it will not germinate given that the seed is of type A3,
(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.

Answers (1)

Solution

Given-

A1, A2, and A3 denote the three types of flower seeds and A1: A2: A3 = 4: 4 : 2
Therefore, Total outcomes = 10

Therefore, \mathrm{P}\left(\mathrm{A}_{1}\right)=\frac{4}{10}, \mathrm{P}\left(\mathrm{A}_{2}\right)=\frac{4}{10} \text { and } \mathrm{P}\left(\mathrm{A}_{3}\right)=\frac{2}{10}

Let E be the event that a seed germinates and E’ be the event that a seed does not germinate.
P(E|A1) is the probability that the seed germinates when it is seed A1.
P(E’|A1) is the probability that the seed will not germinate when it is seed A1.
P(E|A2) is the probability that seed germinates when it is seed A2.
P(E’|A2) is the probability that the seed will not germinate when it is seed A2.
P(E|A3) is the probability that the seed germinates when it is seed A3.
P(E’|A3) is the probability that the seed will not germinate when it is seed A3.

\\ \therefore P\left(E \mid A_{1}\right)=\frac{45}{100}, P\left(E \mid A_{2}\right)=\frac{60}{100} \text { and } P\left(E \mid A_{3}\right)=\frac{35}{100} \quad \text { and } \\ P\left(E^{\prime} \mid A_{1}\right)=\frac{55}{100}, P\left(E^{\prime} \mid A_{2}\right)=\frac{40}{100} \text { and } P\left(E^{\prime} \mid A_{3}\right)=\frac{65}{100}

The Law of Total Probability:

In a sample space \mathrm{S},$ let $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \ldots \ldots .$ En be $\mathrm{n} mutually exclusive and exhaustive events are associated with a random experiment. If A is any event which occurs with \mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \ldots \ldots$ $\mathrm{E}_{0,} then \left.\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)+\ldots \ldots \mathrm{P}\left(\mathrm{E}_{n}\right) \mathrm{P}(\mathrm{A}] \mathrm{E}_{n}\right)

(i) Probability of a randomly chosen seed to germinate.

It can be either seed A, B or C, therefore,

From the law of total probability,

\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{A}_{1}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{A}_{1}\right)+\mathrm{P}\left(\mathrm{A}_{2}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{A}_{2}\right)+\mathrm{P}\left(\mathrm{A}_{3}\right) \times \mathrm{P}\left(\mathrm{E} \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{180}{1000}+\frac{240}{1000}+\frac{70}{1000} \\\\=\frac{490}{1000} \\\\=0.49

(ii) that it will not germinate given that the seed is of type A3

Knowing that P(A) + P(A’) =1

P(E’|A3) = 1 – P(E|A3)

\\ =1-\frac{35}{100} \\ =\frac{65}{100}

(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

\\ \therefore \mathrm{P}(\mathbf{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathbf{B} \mid \mathrm{A})}{\mathrm{P}(\mathrm{B})} \\ \mathrm{P}\left(\mathrm{A}_{2} \mid \mathrm{E}^{\prime}\right)=\frac{\mathrm{P}\left(\mathrm{A}_{2}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{2}\right)}{\mathrm{P}\left(\mathrm{A}_{1}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{1}\right)+\mathrm{P}\left(\mathrm{A}_{2}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{2}\right)+\mathrm{P}\left(\mathrm{A}_{3}\right) \times \mathrm{P}\left(\mathrm{E}^{\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{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}} \\ =\frac{160}{510} \\ =\frac{16}{51}

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