## mbzruby one year 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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1. kirbykirby

Method 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 $P(S \cup M) = P(S) + P(M) - P(S \cap M)\\ \implies P(S \cap M)=P(S) + P(M) - P(S \cup M) =0.44+0.78-0.85=0.37$ Back to the above equation, we can now solve for it: $P(S\cap \overline{M})=0.44-0.37=0.07 :)$ --------------- 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 $(0.44-x)+x+(0.78 - x) = 0.85 \\ \implies x = 0.37$ 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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