Should we advertise?

Probability Level 3

Sanjoy's imaginary billion dollar company advertises only in magazines. An analysis made by his marketing team shows that of all potential customers, only 20% buy their products. Of the customers that buy, exactly 35% have seen the ads. Of the customers that don’t buy, exactly 40% have seen the ads. To the nearest hundredth, what is the conditional probability that those who have seen the ads will buy the product?

The answer is 0.18.

5 solutions

Aditya Joshi
Feb 23, 2014

Bayes' rule

The first thing to do when you see an application of bayes rule is to write down what you want to find. Let $B$ denote the event that the customer buys the product. Let $S$ denote the event that the customer does not buy the product.

We want to find $P(B | S)$

Thus, by Bayes' rule,

$P(B|S) = \dfrac{P(S|B) \times P(B)}{P(S)}$ and

$P(S | B) = \dfrac{35}{100}$ and $P(B) = \dfrac{20}{100}$ also,

$P(S) = P(S | B) \times P(B) + P(S | \neg B) \times P(\neg B)$

Thus, $P(S) = \dfrac{35}{100} \times \dfrac{20}{100} + \dfrac{40}{100} \times \dfrac{80}{100}$

Our final answer is $\dfrac{0.35 \times 0.20}{0.35 \times 0.20 + 0.40 \times 0.80} = \boxed{0.17948717}$ or $\boxed{0.18}$

Sina Soltanieh
Feb 24, 2014

Yay, i got it! 20% of .35 is .07 80% of .4 is .32 multiply by 100- so you have 7 and 32. Then you do 7/(7+32) or 7/39 which is .17948717, or to the nearest hundredth, .18

this is the easiest means. You assume 100 customers and based on given percentage find the answer as explained by Sina. :)

Sushil Bharati - 7 years, 2 months ago

Same!!!!!!!!

Krishna Ar - 6 years, 11 months ago

Kartik Prabhu
Jul 8, 2014

Isn't this question quite simple really? For the sake of nice numbers, lets say that there are 100 people. Then

$\frac{80}{100} \times \frac{40}{100} = \frac{32}{100}$

people see the ad and don't buy anything.

Also:

$\frac{35}{100} \times \frac{20}{100} = \frac{7}{100}$ ,

which are the people who see the ad and buy the product.

That means in total, $7 + 32 = 39$ people out of the 100 see the ad, and just $7$ of those people buy the product.

So the conditional probability $= \frac{7}{39} = \boxed{0.17948}$

Rajesh Ch
Mar 11, 2014

20% buy things . out of 20%, 7% seen the ads. since 20 35/100=7 and 80% does not buy things out of 80%,32% only see ads. since 80 40/100=32 then required answer is 7/(7+32)=0.17948

Jais George
Feb 18, 2014

No.of potential customers = x No.of customers who buy be y. No.of customers who buy seeing the ad be z. No.of customers who don’t buy seeing the ad be a.

Given: y = x/5. z = 35%y = 7%x. a = 32%x. Total no.of outcomes = 32%x + 7%x = 39%x. Favourable no.of outcomes = 7%x. That gives: Probability = 7%x / 39%x = 7 / 39. That’s the answer. THANK YOU...

The problem should specify whether the answer should be in percent or decimal form.

Kenneth Wang - 7 years, 3 months ago

the maximum probability of an event is "1". Probability is always expressed in fraction not in percentage terms.

Ln Jena - 7 years, 3 months ago

ya i also have solved it upto .17 n my solution indicated wrong.

MOHD SHAHNAWAZ QURESHI - 7 years, 3 months ago

Note that you cannot type percentages into the answer field. 18% = 0.18.

Calvin Lin Staff - 7 years, 3 months ago

Sorry, to hijack your solution with a comment, but I think a bit of cleaner explanation will help (and I accidentally made my solution private before even finishing it, whoops.) This is also a great venture (and introduction, I'm sure, for some) into conditional probabilities.

Let $A$ represent the event of a potential customer seeing an advertisement, and let $B$ represent the event of a potential customer buying the product.

Let $P(X|Y)$ represent the conditional probability of event $X$ occurring given that event $Y$ occurs, and let $P(X')=1-P(X)$ represent the probability of event $X$ not occurring.

Then the question gives us the following: $\begin{aligned}&P(B)=0.2,\\&P(A|B)=0.35,\\&P(A|B')=0.4\end{aligned}$ It asks us to find $P(B|A)$ .

Firstly, we will find $P(A)$ . Note that $P(A)=P(A\cap B)+P(A\cap B')$ . In addition, $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ , so $P(A\cap B) = P(A|B)P(B)$ . Therefore, $\begin{aligned}P(A)&=P(A\cap B)+P(A\cap B')\\&=P(A|B)P(B)+P(A|B')P(B')\\&=(0.35)(0.2)+(0.4)(1-P(B))\\&=(0.35)(0.2)+(0.4)(1-0.2)\\&=0.39\end{aligned}$

We can now calculate $P(B|A)$ . Using Bayes' theorem of conditional probabilities, $\begin{aligned}P(B|A)&=\frac{P(A|B)P(B)}{P(A)}\\&=\frac{(0.35)(0.2)}{0.39}\\&=\frac{7}{39}\\&\approx\boxed{0.18}\end{aligned}$

Jonathan Wong - 7 years, 3 months ago

hey idid exactly the same but mine came to be wrong. only mistake was i solved upto .17 gosh i was so close.

Harman Deep - 7 years, 3 months ago

Should we advertise?

The answer is 0.18.

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