Laplace's rule

Laplace's rule is extremely important because it allows us to compute the probability of an event, always that the elementary events are equiprobable, that is, that all possible outcomes have the same probability. Under these conditions, we have:

  • The probability of an event $$A$$ is obtained by dividing the number of results that form the event $$A$$ by the number of possible outcomes.

If we say that the events of $$A$$ are the favorable cases of $$A$$, then we can write the Laplace's rule as:

$$$P(A)=\dfrac{\mbox{favorable cases to } A}{\mbox{possible cases}} $$$

Attention! We should take into account that this rule only works when all cases are equiprobable.

  • It is NOT valid when we calculate the probability of event $$A=$$"being hit by tram". There are only two possible cases, "run over" and "not get hit". There is a favorable case, "being hit". Therefore, the probability is $$P(A)=\dfrac{1}{2}$$.

This would mean that every time we get out to the street, we would have a $$50\%$$ chance of being hit. The mistake is, of course, that the two events do not have the same probability.

By the same reason, if we solve an exercise with Laplace's rule, we must be careful to ensure that our possible results are not mixed up since they might not all be equally probable.

Consider the following example NOT valid:

If a family has two children, and we assume that the probability of being a man is the same as that of being a woman, what is the probability that the two children are of the same sex?

We would argue that the sample space is "two men", "a man and a woman" and "two women." Then, the favorable cases are $$2$$, and $$3$$ cases are possible, so that the probability would be $$2/3$$.

The error is that the three cases are not equally probable since it could turn out that the first child turns out to be a man, and the second one a woman, or that the first one is a woman, and the second one, a man. We will solve it properly in the following example.

So, when we want to apply the Laplace's rule, we are used to consider our sample space in order, as we have done so far, in order to not make errors.

Let's see how to apply this rule properly:

If a family has two children, and we assume that the probability of being a man is the same as that of being a woman, what is the probability of having both children of the same sex?

As we are going to apply the Laplace's rule, we will consider the results in order.

In this case, our sample space is $$\Omega=\{HH,HM,MH,MM\}$$.

Favorable cases are $$HH$$ and $$MM$$. Therefore, the probability is $$$\dfrac{2}{4}=\dfrac{1}{2}$$$ which is $$0'5$$ (a $$50\%$$).

Calculate the probability of throwing two coins and getting Heads both times.

Every time you throw a coin it is equally probable to come out with heads or tails, so that all possible outcomes are equiprobable.

Our sample space has four elements $$\Omega=\{hh,ht,th,tt\} $$, and there is only one case in favor of event $$A=$$"get heads twice"$$=\{hh\}$$.

So, it is a probability of $$$P(A)=\dfrac{1}{4}$$$ which is $$0'25$$, a $$25\%$$.

Suppose that the probability of being born male is the same as the probability of being born female. What is the probability that, if we have three children, two of them are women?

Our elementary events will be: "man" $$(H)$$ and "woman" $$(M)$$. Our sample space is then, if we consider the children sorted $$\Omega=\{HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM\}$$.

Then, the favorable cases to our event $$A=$$"two women" are: $$HMM, MHM, MMH, MMM$$.

So, $$$P(A)=\dfrac{4}{8}=\dfrac{1}{2}$$$ that is, we have a probability of $$0'5$$, i.e., $$50\%$$.

It is correct also to assume that the probability of each event is $$\dfrac{1}{8}$$, as they are all equiprobable (and the sum of all probabilities must be 1), and therefore, if we have four favorable elementary events

$$$ P(A)=P(HMM)+P(MHM)+P(MMH)+P(MMM)=4\cdot\dfrac{1}{8}=\dfrac{1}{2}.$$$

Suppose we throw a dice. We calculate the probability of $$A =$$ "get an even number, $$B =$$" to get a number less than or equal to $$5$$", $$C =$$" to get a three or a five ", $$D=A\cap B$$, $$E=A\cup B$$.

The sample space is $$\Omega=\{1,2,3,4,5,6\}$$. Favorable cases to $$A: 2,4,6$$. Favorable cases to $$B: 1,2,3,4,5$$. Favorable cases to $$C: 3,5$$.

Therefore, $$P(A)=\dfrac{3}{6}=\dfrac{1}{2}$$, $$P(B)=\dfrac{5}{6}$$, $$P(C)=\dfrac{2}{6}=\dfrac{1}{3}.$$

Now let's calculate that $$D =$$"get an even number less than or equal to $$5$$". Favorable cases are: $$2, 4$$. Therefore, $$P(D)=\dfrac{2}{6}=\dfrac{1}{3}.$$

For its part, $$E=$$"get an even number, or less than or equal to $$5$$"$$= \{1,2,3,4,5,6\}$$. Therefore, $$P(E)=\dfrac{6}{6}=1.$$

Note that it is true that: $$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

This is a general property of probability.

In our case, as $$A\cup B$$ it's all the sample space and $$A\cap B=D$$, we have $$P(\Omega)=P(A)+P(B)-P(D)$$, that is, $$$1=\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{3+5-2}{6}=\dfrac{6}{6}$$$