anonymous
  • anonymous
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.
Probability
jamiebookeater
  • jamiebookeater
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kirbykirby
  • 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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