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A shopkepper sells three types of flower...

A shopkepper sells three types of flower seeds `A_(1),A_(2) and A_(3)`. They are sold as mixture, where the proportion are 4 : 4 : 2, respectively. The germination rates of the three types of seeds 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 bot germinate

Text Solution

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W have, `A_(1) : A_(2) : A_(3)=4 : 4 : 2`
`P(A_(1))=4/10,P(A_(2))=4/10and P(A_(3))=2/10`
where `A_(1),A_(2) and A_(3)` denote the three types of flower seeds.
Let E be the event that aseeed germinbates and `barE` be the event that a seed does not germinate.
`thereforeP(E//A_(1))=45/100,P(E//A_(2))=60/100and P(E/A_(3))=35/100`
and `P(barE//A_(1))=55/100,P(barE//A_(2))=40/100and P(barE/A_(3))=65/100`
(i) `therefore P(E)=P(A_(1))cdotP(E//A_(1))+P(A_(2))cdotP(E//A_(2))+P(A_(3))cdotP(E/A_(3))`
`=4/10cdot45/100+4/10cdot60/100+2/10cdot35/100`
`=180/1000+240/1000+70/1000=490/1000=0.49`
(ii) `P(barE//A_(3))=1-P(E//A_(3))=1-P(E//A_(3))=1-35/100=65/100` [as given above]
(iii) `P(A_(2)/barE)=(P(A_(2))cdotP(barE//A_(2)))/(P(A_(1))cdotP(barE//A_(1))+P(A_(2))cdotP(barE//A_(2))+P(A_(3))cdotP(barE//A_(3)))`
`=(4/10cdot40/100)/(4/10cdot55/100+4/10cdot40/100+2/10cdot65/100)=(160/1000)/(220/1000+160/1000+130/1000)`
`=(160//1000)/(510//1000)=16/51=0.313725=0.314`
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