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Consider the following information about travelers on vacation: $40$% check work email, $30$% use a cell phone, $25$% bring a laptop with them, $23$% both check work email and use a cell phone, and $51$% neither check work email nor use a cell phone nor bring a laptop. In addition $88/100$ who bring a laptop also check work email, and $70/100$ who use a cell phone also bring a laptop.

What is the probability that someone who brings a laptop on vacation also uses a cell phone?

I let $A$ represent check work email, $B$ represent use cell phone, and $C$ represent brings laptop. I have the following probabilities:

$$\begin{align}P(A)&=.4, \\P(B)&=.3, \\P(C)&=.25,\\ P(A\cap B)&=.23, \\P(A\cup B\cup C)&=.49,\\ P(A \mid C)&=.88, \\P(C \mid B)&=.7\end{align}$$

I found a solution online saying to use Bayes' Theorem to solve this part of the problem but I do not understand why to use Bayes' Theorem. Here is a picture to the solution I am referring to: Problem

Bayes' Theorem does not look like what the solution says to use. In my textbook, the theorem looks like this:Bayes Theorem

Did the person that wrote the solution simplify something? It does not look like either of these two forms.

Also, for problems like these, is there a general rule on when to use Bayes' Theorem and the rule for Total Probability? I cannot figure out when to use what.

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1 Answer 1

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The law of total probability is used in Bayes theorem:

$P(A|B)=\frac{P(A\cap B)}{P(B)} \implies P(A\cap B) = P(B)P(A|B).$ This is just the definition of conditional probability.

Now, the Law of Total Probabiliyy can be used to calculate $P(B)$ in the above definition. The law requires that you have a set of disjoint events $D_i$ that collectively "cover" the event $B$. Then, instead of calculating $P(B)$ directly, you add up the intersection of $B$ with each of the events $E_i$:

$P(B)=\sum P(B\cap E_i)$ Of course, we can rewrite this using the definition of conditional probability:

$P(B)=\sum P(B\cap E_i)=\sum P(E_i)P(B|E_i)$

Thus, the following are equivalent:

$P(B|A)= \frac{P(A\cap B)}{P(A)} = \frac{P(B)P(A|B)}{P(B)P(A|B)+P(\neg B)P(A|\neg B)}$ Since $B$ and $\neg B$ are disjoint events.

In general, Bayes' rule is used to "flip" a conditional probability, while the law of total probability is used when you don't know the probability of an event, but you know its occurrence under several disjoint scenarios and the probability of each scenario.

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  • $\begingroup$ In the solution that I provided, the denominator has $P(C)$, could you explain why this is? $\endgroup$
    – Kot
    Commented Sep 28, 2014 at 4:54
  • $\begingroup$ @Kot: Yes, because you are trying to find the proportion of computer users that use cell phones. Therefore, the fraction of the total population that uses computers is $P(C)$, and the fraction of the total population that use cell phones and laptops is $P(B \cap C)$. $\endgroup$
    – user76844
    Commented Sep 28, 2014 at 17:03
  • $\begingroup$ @Kot the fact that you are confused by variants in form suggests that you should re-visit the basic concept of what conditional probability is, so you are not dependent on any particular mathematical form for its expression. The essence of conditional probability is simply that you are changing what you consider the "sample space" from "all possible events" to "events that satisfy condition C". Since $P(C)\leq 1$, you need to "renormalize" by dividing by $P(C)$ so that the total probability in the new sample space sums to one. $\endgroup$
    – user76844
    Commented Sep 28, 2014 at 17:07
  • $\begingroup$ From what you're saying, the denominator should be the entire sample space of the given condition, in this case $C$? $\endgroup$
    – Kot
    Commented Sep 28, 2014 at 21:21
  • $\begingroup$ @Kot exactly! Only events that have an intersection with $C$ can be in the sample space. $\endgroup$
    – user76844
    Commented Sep 28, 2014 at 23:19

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