If given two events $E$ and $F$ are given, which are mutually exclusive, then the textbook showed that the probability of $E$ being occured before $F$ occurs is $$\frac{P(E)}{P(E)+P(F)}$$ Its proof was, since there is no $F$ happening before $E$, only $(E\cap F)^c$ should occur before $E$ and hence $$\sum_{n=1}^\infty {P\left([E\cap F]^c\right)^nP(E)} = \frac{P(E)}{1-P\left([E\cap F]^c\right)} = \frac{P(E)}{P(E)+P(F)}$$ and $P(EF)=0$ is used on the last equality. Now I would like to apply this to the following problem.
If a pair of dice is rolled in each trial, let event $E$ be the situation of happening 6 times of the sum being even and $F$ be the one of occuring 2times of the sum being 7. Find the probability that $F$ happens before $E$.
My consideration was to directly subsitute to the above formula, but then it remained as a doubt that $E$ and $F$ might not be exclusive. Is it possible to directly put them in the equation or is something should be modified? Appreciate with this.
For what it's worth, the description excerpted above suggests to me that the corresponding Probability text book is poorly written. The above presentation is cart before the horse.
Before you dump a valid (but highly abstract) theorem on the student, it is a good idea to first (greatly) expand the student's intuition with several unrelated examples.
You would then have each example seem sensible to the student. Then, you would present the theorem as a unifying idea, with the examples being applications of the theorem. So, before the student is exposed to either the statement of the theorem, or the proof to the theorem, the student's intuition has very gracefully been stretched.
I guess not all Mathematicians are good at writing Math books.
Assume that all dice rolls that are odd but not equal to $7$ are ignored. Assume that you roll the dice over and over again, until you get exactly $7$ occurrences of the sum of the dice being either even or equal to $7$.
Since you are rolling exactly $7$ times, not counting the rejected dice rolls, one of $2$ things will happen:
Note that the two possibilities above are both disjoint and comprehensive (i.e. exactly one of the two possibilities must occur).
In attacking this problem, the first step is to determine the relative probabilities of a single roll of the dice being $7$ or even.
Of the $36$ possible dice rolls, $6$ of them will result in a $7$, while $18$ of them will result in an even roll.
So, the relative probabilities are $(1/6)$ versus $(1/2)$.
Therefore, on any dice roll where the odd rolls not equal to $7$ are ignored, the probabilities are:
At this point, the problem reduces to a Binomial Distribution, specifically $\displaystyle \binom{n}{k}p^kq^{(n-k)}.$
With $p = (1/4)$ and $q = (1-p)$, the probability of getting either $0$ occurrences of a $7$ or $1$ occurrence of a $7$ (where odd rolls $\neq 7$ are ignored) is
$$\binom{7}{0}p^0q^7 + \binom{7}{1}p^1q^6.$$
Taking out the common factor of $\frac{1}{4^7}$, you have that the probability of having exactly $0$ or $1$ rolls of a $7$, out of $7$ dice rolls is
$$\frac{1}{4^7} \times [(1 \times 3^7) + (7 \times 3^6)].$$
This may be more simply expressed as
$$\frac{3^6}{4^7} \times (3 + 7).$$