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In order to gauge the effectiveness of the OHaganBooks.com site, you recently commissioned a survey of online shoppers. According to the results, 2% of online shoppers visited OHaganBooks.com during a one-week period, while 5% of them visited at least one of OHaganBooks.com’s two main competitors: JungleBooks.com and FarmerBooks.com. Use this information to answer. Assuming that visiting OHaganBooks.com was independent of visiting a competitor, what percentage of online shoppers visited either OHaganBooks.com or a competitor?

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Answered 1 year ago
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Denote the events as follows:

H={ a r a n d o m l y c h o s e n o n l i n e s h o p p e r v i s i t e d O H a g a n B o o k s . c o m }C={ a r a n d o m l y c h o s e n o n l i n e s h o p p e r v i s i t e d e i t h e r o f t h e m a i n c o m p e t i t o r s }\begin{align*} & H = \qty{\text{a randomly chosen online shopper visited OHaganBooks.com}}\\ & C = \qty{\text{a randomly chosen online shopper visited either of the main competitors}} \end{align*}

We are given the following information:

P(H)=2%=0.02P(C)=5%=0.05\begin{align*} & P(H) = 2\% = 0.02\\ & P(C) = 5\% = 0.05 \end{align*}

Also, HH and CC are said to be independent which means that P(HC)=P(H)P(C)P(H\cap C) = P(H)P(C).

We want to find the percentage of online shoppers that visited either OHaganBooks.com or a competitor i.e. we want to find P(HC)P(H\cup C). Using the addition principle for probabilities, we get

P(HC)=P(H)+P(C)P(HC)=P(H)+P(C)P(H)P(C)=0.02+0.050.02×0.05=0.069=6.9%\begin{align*} P(H\cup C) &= P(H)+P(C)-P(H\cap C)\\ &= P(H)+P(C)-P(H)P(C)\\ &= 0.02+0.05-0.02\times0.05\\ &= 0.069\\ &=6.9\% \end{align*}

Therefore, 6.9% of online shoppers visited either OHaganBooks.com or a competitor.

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