## Solution :

Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2

\(\therefore\) P(A) + P(B) + P(C) = 1

then events A, B, C are exhaustive.

If P(E) = Probability of introducing a new product, then as given

P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5

= 0.5 \(\times\) 0.7 + 0.3 \(\times\) 0.6 + 0.2 \(\times\) 0.5 = 0.35 + 0.18 + 0.10 = 0.63

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