If a consumer is selected from the survey at random, how likely is it that he/sh

Question

If a consumer is selected from the survey at random, how likely is it that he/she either saw the ad or purchased the product or did both? (Round answer to 2 decimal places, e.g. 52.22.)

Section Exercise 4.49 A questionnaire concerning the effect of a recent advertising campaign was sent out to a sample of 570 consumers. Results of the study are reported in the cross-tabulation table below: Saw Didn't Total Ad See Ad 130 Purchased Didn't Purchase 100 Total 50 180 290 390 340 570 230 (al) Your answer is correct. Convert the table to a joint probability table. (Round answers to 2 decimal places, e.g. 52.22.) Saw Ad Didn't See Ad Total Purchased 23 09 32 Didn't Purchase 18 51 68 Total 40 60

Explanation / Answer

Ans:

P(saw ad or purchased)=P(saw ad)+P(purchase)-P(saw ad and purchased)

=0.4+0.32-0.23=0.49

P(saw ad or purchased) =0.49

P(saw ad and purchased)=0.23

Now,if you need probability that saw the ad or purchased the product or did both together,it will be simply

P(saw ad or purchased)+P(saw ad and purchased)=0.49+0.23=0.72

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