A) 0.18
B) 0.35
C) 0.10
D) 0.63
Correct Answer: D
Solution :
Let \[E\] be the event that a new product is introduced. Then \[P(A)=0.5\], \[P(B)=0.3\], \[P(C)=0.2\] and \[P(E/A)=0.7,\,\,P(E/B)=0.6,\,\,P(E/C)=0.5.\] \[\because \,\,\,A,\,\,B\] and \[C\] are mutually exclusive and exhaustive events. \[P(E)=P(A)\,.\,P(E/A)+P(B)\,.\,P(E/B)+P(C)\,.\,P(E/C)\] \[=0.5\times 0.7+0.3\times 0.6+0.2\times 0.5\] \[=0.35+0.18+0.10=0.63.\]You need to login to perform this action.
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