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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

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Question

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

Sum
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Solution

Given that A1: A2: A3 = 4: 4: 2

∴ P(A1) = `4/10`

P(A2) = `4/10`

And P(A3) = `2/10`

Where A1, A2 and A3 are the three types of seeds.

Let E be the event that a seed germinates and `bar"E"` be the event that a seed does not germinate

∴ `"P"("E"/"A"_1) = 45/100 "P"("E"/"A"_2) = 60/100` and `"P"("E"/"A"_3) = 35/100`

And `"P"(bar"E"/"A"_1) = 55/100, "P"(bar"E"/"A"_2) = 40/100` and `"P"(bar"E"/"A"_3) = 65/100`

Using Bayes’ Theorem, we get

`"P"("A"_2/bar"E") = ("P"("A"_2)*"P"(bar"E"/"A"_2))/("P"("A"_1)*"P"(bar"E"/"A"_1) + "P"("A"_2)*"P"(bar"E"/"A"_2) + "P"("A"_3)*"P"(bar"E"/"A"_3))`

= `(4/10*40/100)/(4/10*55/100 + 4/10*40/100 + 2/10*65/100)`

= `(160/1000)/(220/1000 + 160/1000 + 130/1000)`

= `160/(220 + 160 + 130)`

= `160/510`

= `16/51`

= 0.314

Hence, the required probability is `16/51` or 0.314

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Chapter 13: Probability - Exercise [Page 276]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 13 Probability
Exercise | Q 43.(iii) | Page 276

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