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.

Here, A1, A2, and A3 denote the three types of flower seeds.

and A1: A2: A3 = 4: 4 : 2


Total outcomes = 10


So,


Let E be the event that a seed germinates and E’ be the event that


a seed does not germinate.


Now P(E|A1) is the probability that seed germinates when it is seed A1.


P(E’|A1) is the probability that 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 seed will not germinate when it is seed A2.


P(E|A3) is the probability that seed germinates when it is seed A3.


P(E’|A3) is the probability that seed will not germinate when it is seed A3.



and



The Law of Total Probability:


In a sample space S, let E1, E2, E3……. En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1, E2, E3……. En, then


P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + …… P(En)P(A|En)


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


It can be either seed A, B or C.


So,


From law of total probability,


P(E) = P(A1) × P(E|A1) + P(A2) × P(E|A2) + P(A3) × P(E|A3)





= 0.49


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


As we know P(A) + P(A’) =1


P(E’|A3) = 1 – P(E|A3)




(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.


We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.








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