A new marketing campaign for an existing product that currently has 5% of the market runs advertisements in newspapers. Based on circulation and previous survey information, the marketing company states that the probability an arbitrary person sees the ads is 0.6. During the newspaper ads campaign, the buyers of the product were asked if they’d seen the ads – 75% replied that they had. Assume that for those who did not see the ads, the probability of buying the product is still 0.05.Show that the probability that a person who saw the ads buys the product is 0.1.
[Math] Probability Saw the ads buy the product
probabilityprobability theory
Related Solutions
Let $A$ be the case where the observer saw the advertisement, and let $B$ represent the case that the observer bough the item.
What we know is the following:
- $P(B|A) = .56$, the probability that an observer buys the item given that they saw the ad is 56%
- $P(B|\overline{A}) = .08$, the probability that an observer buys the item given that they did not see the ad is 8%
- $P(A) = .25$, the probability that someone saw the ad, regardless of whether they bought an item or not is 25%.
The questions are to find a) $P(B)$ and b) given 5 people, what is the probability that at least one will buy the product.
We must solve a) before we can address b). Even though you have the correct answer, I'm not sure what you meant when you described how you got it, so for completeness, I will post a derivation below.
Question A
This is a case of conditional probability, which is almost always handled by Bayes' Rule.
$P(B)$ is made up of two cases:
- $P(B \cap A)$, the probability that someone who saw the advertisement bought the item.
- $P(B \cap \overline{A})$, the probability that someone who did not see the advertisement bought the item.
From Bayes Rule, we know that: $$ P(B|A) = \frac{P(B \cap A)}{P(A)}\\ P(B|\overline{A}) = \frac{P(B \cap \overline{A})}{P(\overline{A})} $$
Starting with the first case, we know both $P(B|A)$ and $P(A)$, so solving for $P(B \cap A)$ gives us: $$ .56 = \frac{P(B \cap A)}{.25}\\ P(B \cap A) = 0.14 $$ We address $P(B \cap \overline{A})$ similarly. Given that everyone has to either see or not see the ad, we know that $P(\overline{A}) = 1 - P(A) = 0.75$. So $$ .08 = \frac{P(B \cap \overline{A})}{.75}\\ P(B \cap \overline{A}) = 0.06 $$
Now, $P(B)$ comprises everyone who bought the item whether or not they saw the ad. Symbolically, this means $P(B) = P(B \cap A) + P(B \cap \overline{A})$, as the two possibilities are disjoint. We know that now, that is $.14 + .06 = .2$ as you said.
Question B
Any one person buying the item or not is an example of a Bernoulli random variable, a random variable which can only take one of two values (e.g. Yes/No, Heads/Tails, Bought/Did not buy). You are asked a question about a group of Bernoulli variables (5 people). The mathematics that describes the probabilities of a group of independent Bernoulli random variables is the binomial distribution. The other point of note is that you are asked for the probability that "at least one person" buys an item. That is all possible cases except for the case where no one buys the item. So we have to figure out the probability that out of 5 people, no one buys the item, when each individual person has a 20% chance of buying the item. The appropriate formula is: $$ {5 \choose 0}{0.2}^0{0.8}^5 = 1 \times 1 \times \frac{1024}{3125} $$
So the answer to b) is: $$ 1 - \frac{1024}{3125} = \frac{2101}{3125} = .67232 $$
Let event $A$ be owning an automobile. Let event $B$ be owning a house.
Consider the diagram below:
If we simply add the probabilities that a person owns an automobile and a person owns a house, we will have added the probability that a person owns both an automobile and a house twice. Thus, to find the probability that a person owns an automobile or a house, we must subtract the probability that a person owns both an automobile and a house from the sum of the probabilities that a person owns an automobile and that a person owns a house.
$$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$
Note that on the left-hand side of your equation, you should have written $\Pr(A \cup B)$ or $\Pr(A~\text{or}~B)$ rather than $\Pr(A)~\text{or}~\Pr(B)$.
Since we are given that $60\%$ of the population owns an automobile, $30\%$ of the populations owns a house, and $20\%$ owns both, $\Pr(A) = 0.60$, $\Pr(B) = 0.30$, and $\Pr(A \cap B) = 0.20$. Hence, the probability that a person owns an automobile or a house is
$$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$$
However, the question asks for the probability that a person owns an automobile or a house, but not both. That means we must subtract the probability that a person owns an automobile and a house from the probability that the person owns an automobile or a house.
$$\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B) = 0.70 - 0.20 = 0.50$$
In terms of the Venn diagram, $A \cup B$ is the region enclosed by the two circles, while $A~\triangle~B = (A \cup B) - (A \cap B) = (A - B) \cup (B - A)$ is the region enclosed by the two circles except the region where the sets intersect.
Since $A - B = A - (A \cap B)$,
$$\Pr(A - B) = \Pr(A) - \Pr(A \cap B) = 0.60 - 0.20 = 0.40$$
Since $B - A = B - (A \cap B)$,
$$\Pr(B - A) = \Pr(B) - \Pr(A \cap B) = 0.30 - 0.20 = 0.10$$
Hence,
$$\Pr(A~\triangle~B) = \Pr(A - B) + \Pr(B - A) = 0.40 + 0.10 = 0.50$$
which agrees with the result obtained above.
Best Answer
Let $p$ be the probability that a person who sees the ad buys the product. Assuming that the marketing company is correct, the probability that a randomly chosen person buys the product is $$(0.4)(0.05)+(0.6)(p).$$ For $40\%$ of all people have not seen the ad, and they have probability $0.05$ of buying, while $60\%$ have seen the ad, and they have probability $p$ of buying.
Thus the probability that a person has seen the ad, given that she buys the product, is $$\frac{(0.6)(p)}{(0.4)(0.05)+(0.6)(p)}.$$ We are told this is $0.75$. So set the above expression equal to $0.75$, and solve for $p$. Or else, note that there is no more than $1$ solution. so plug in $0.1$ fpr $p$, and verify that we indeed get $0.75$.