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mbzruby
 8 months ago
What is the probability that the person thinks that "Made in America" ad boosts sales but does not use social media? 44% thinks adds boost sales, 78% use online media, 85% thinks boost sales or use social media online.
mbzruby
 8 months ago
What is the probability that the person thinks that "Made in America" ad boosts sales but does not use social media? 44% thinks adds boost sales, 78% use online media, 85% thinks boost sales or use social media online.

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kirbykirby
 8 months ago
Best ResponseYou've already chosen the best response.1Method 1: Algebraically Say \(S\) = think ads boost sales \(M\) = use online media You are given: \(P(S)=0.44\\ P(M)=0.78 \\ P(S \cup M) = 0.85\) You are interested in: \(P(S \cap \overline{M})\) Notice that:\[P(S)=P(S\cap M)+P(S \cap \overline{M})\\\implies P(S \cap \overline{M})=P(S)  P(S \cap M) \] and \[ P(S \cup M) = P(S) + P(M)  P(S \cap M)\\ \implies P(S \cap M)=P(S) + P(M)  P(S \cup M) =0.44+0.780.85=0.37\] Back to the above equation, we can now solve for it: \[ P(S\cap \overline{M})=0.440.37=0.07 :)\]  Method 2: Venn diagram dw:1408192137497:dw If you set the intersection to be \(x\), then the region that is just (S and not M) is \(0.44  x\), and the region that is just (M and not S) is \(0.78  x\). Since we know that P(S or M) = \(P(S \cup M)\) in the problem is 0.85, then \[ (0.44x)+x+(0.78  x) = 0.85 \\ \implies x = 0.37\] Now, since the region of interest was (S and not M) = \(S \cap \overline{M}\) which was \(0.44 x\), then this is \(0.44  0.37 =0.07 :) \)
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