Probability of a customer seeing the advertisement in at least one of two media?

[kshah93]

Homework Statement

A computer store advertised its annual half-price sale in the newspaper and on TV. A survey of 200 customers indicated that 60 learned about the sale from the newspaper, 50 from TV, and 30 from both sources. What is the probability of the following events?
b) A randomly selected customer saw the advertisement in at least one of the two media.

Homework Equations

P(A or B)= P(A) + P(B) - P(A and B)

The Attempt at a Solution

Let A be Newspaper
Let B be TV

P(A or B) = (60 + 30 + 50) / 200 = 7/10

Apparently my answer is wrong. The correct answer in the back of my textbook is 2/5. I know how to calculate this using the equation I provided in the relevant equation section, but I don't see why I would use this. My confusion lies in the fact that it says that 60 out of 200 customers learned from the newspaper, 50 out of 200 learned from the TV, and 30 out of 200 learned from both. So why would you use the equation above when each event is mutually exclusive. I mean it says that 60 learned from the TV. That means that 60 out of 200 only learned from the TV. There is no intersection between these events. Can someone help me understand this problem?

 

[Square]

It should be

P(A or B) = (60+50-30)/200 = 80/200 = 2/5

 

[kshah93]

It should be

P(A or B) = (60+50-30)/200 = 80/200 = 2/5

Yeah, I know that but why? I mean it gives you the total number of customers (200) and it tells you 60 out of those 200 learned from the newspaper, 30 out of 200 learned from both sources and 50 out of 200 learned from TV. So wouldn't you just add all the individual probabilities and divide that by your total number of people? I mean how do we know that the 60 people that learned from the newspaper are completely different from those that learned from the newspaper and those that learned from both?

 

[Square]

Yeah, I know that but why? I mean it gives you the total number of customers (200) and it tells you 60 out of those 200 learned from the newspaper, 30 out of 200 learned from both sources and 50 out of 200 learned from TV. So wouldn't you just add all the individual probabilities and divide that by your total number of people?

I think you might be right, and the book wrong, since it says "saw the advertisement in at least one of the two media."

Then it should be

1-(60/200) = 7/10

The complement of the probability that the customer did not see the advertisement at all. Which is the same as your result.

 

[ehild]

The probability is the number of favourable possibilities divided by the number of all possibilities to choose a person out of 200. The favourable possibility is that the person saw the advertisement at least at one place. The set of such people is the union of those who saw it in newspapers (A) and those who saw it in TV (B): A+B. This union of the sets can be decomposed into three disjunct sets: A\B,(blue) B\A (yellow) and AB (the intersection of A and B, green)
Calculate the cardinality of these three sets, and add them: You get all people who learned about the sale.

ehild

 

[Ray Vickson]

The number who saw the ad in at least one source = N(paper) + N(TV) - N(both), because when we add N(paper) and N(TV), we are double-counting the people in both. That is, N(both) is part of N(paper) and is part of N(TV).

RGV

 

[kshah93]

The number who saw the ad in at least one source = N(paper) + N(TV) - N(both), because when we add N(paper) and N(TV), we are double-counting the people in both. That is, N(both) is part of N(paper) and is part of N(TV).

RGV

But how do you know that those people are not different? Say the 60 people vs the 50 people vs the 30 people? Are you supposed to assume that the 30 that heard from both sources includes people from the 60 people who heard from the newspaper and the 50 people that heard from the TV? The question is not really clear about this.

 

[kshah93]

The probability is the number of favourable possibilities divided by the number of all possibilities to choose a person out of 200. The favourable possibility is that the person saw the advertisement at least at one place. The set of such people is the union of those who saw it in newspapers (A) and those who saw it in TV (B): A+B. This union of the sets can be decomposed into three disjunct sets: A\B,(blue) B\A (yellow) and AB (the intersection of A and B, green)
Calculate the cardinality of these three sets, and add them: You get all people who learned about the sale.

ehild

I totally agree with you but my question is how do you know that the 30 people that both saw the ad on TV and in the paper include the 50 people that saw it on TV and the 60 people that saw it in the paper. How do you know that the 60 people is not the blue shaded region only, the 30 the green shaded region, and the 50 people the yellow shaded region, giving a total number of (50+30+60) people who saw the ad?

 

[ehild]

Well, the text can be understood in both ways in common speach. But when a mathematician says "60 people learned about the sale from the newspaper" he/she means just that, and does not mean that they learned it only from newspaper. If they meant so they would write "from newspaper exclusively".

ehild

 

[kshah93]

Well, the text can be understood in both ways in common speach. But when a mathematician says "60 people learned about the sale from the newspaper" he/she means just that, and does not mean that they learned it only from newspaper. If they meant so they would write "from newspaper exclusively".

ehild

Thanks. That made things clearer.