A shopkeeper sells three types of flower seeds \(A_1\), \(A_2\), \(A_3\). 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. Calculate the probability in the following cases.
The probability that seed is of type \(A_2\) given that seed germinates.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{24}{49}\)
Let G be the event that the seed germinates. With \(P(A_1) = P(A_2) = 0.4\), \(P(A_3) = 0.2\) and germination rates 0.45, 0.60, 0.35, the total probability of germinating is
\(P(G) = 0.18 + 0.24 + 0.07 = 0.49\)
By Bayes’ theorem:
\( \displaystyle P(A_2 \mid G) = \frac{P(A_2)\,P(G \mid A_2)}{P(G)} = \frac{0.4 \times 0.60}{0.49} \)
\( \displaystyle = \frac{0.24}{0.49} = \frac{24}{49} \)