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A landscaping company has made phone calls to attract clients for the upcoming season. According data, in 15 percent of these calls it got new customers and 80 percent of these new customers had used the services of a competitor in the previous year. It is also estimated that 60 percent of all the people you called used the services of a rival company in the previous year. So it is TRUE that the odds that a call to a person who used the services of a rival company last year will land a new customer is :

a) 0.2 b) 0.1 c) 0.5 d) 0.9 e) 0.4

I tried to do by conditonal probability:

$P(B|A)=0.8∗0.15=0.12$

Where:

$$A = New custumers (15%)

B = New custumers who had used services of a competitor in the previus year (80%)$$

It means 80% given 15%

What Am I doing wrong ?

I guess this is the venn diagram :

enter image description here

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    $\begingroup$ Hi Locker05, welcome to Cross Validated! I'm not able to read the photograph because of the lighting and cropping; it would be helpful to include a clearer diagram. Can you please define your variables ($A$ and $B$) and explain why you thought it should just be the product of those two numbers? (It may help to review the definition of conditional probability.) $\endgroup$ Apr 13, 2021 at 23:19
  • $\begingroup$ @Arya . Thank you. When It say : " . . in 15 percent of these calls it got new customers and 80 percent of these new customers had used the services of a competitor in the previous year. " . A = new costumers, B = costumers who used competitor services "inside" new costumers.. . . . But I'm struggle to understand how to put this inside excersice : "It is also estimated that 60 percent of all the people you called used the services of a rival company in the previous year" $\endgroup$
    – Locker05
    Apr 14, 2021 at 2:40

2 Answers 2

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I'll refresh future readers on the variables you defined.

  • $A$ is whether the callee becomes a new customer.
  • $B$ is whether the callee has used a rival company in the previous year.

The information you've been given is:

  • $P(A) = 0.15$
  • $P(B \mid A) = 0.8$
  • $P(B) = 0.6%$

You tried to compute $P(B \mid A)$ by multiplying $P(B \mid A) \times P(A)$. That's not correct. But $P(B \mid A)$ isn't the quantity you care about anyway. You're asked for

the odds that a call to a person who used the services of a rival company last year will land a new customer

which is $P(A \mid B)$. You have enough information to compute that from the three quantities in the question.


I know the question says 'odds', but none of the answer choices are the correct odds. The correct probability is on the list.

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  • $\begingroup$ So Have I use Bayes Theorem ? Let me try it and I'll tell you. $\endgroup$
    – Locker05
    Apr 18, 2021 at 0:15
  • $\begingroup$ really thank you. I am going to answer my question. By the way : Do you think the venn diagram is well done ? $\endgroup$
    – Locker05
    Apr 18, 2021 at 1:00
  • $\begingroup$ The new one is very clear! $\endgroup$ Apr 19, 2021 at 1:45
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Thank you to Arya.

$P(A) = 0.15$ $P(B) = 0.6$ $P(B|A) = 0.8$

$P(A|B) = \frac{0.15*0.8}{0.6} $ $P(A|B) = 0.2$

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