Consider this question
Consider the experiment of tossing a coin. If the coin shows head, toss it
again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.
Now we know the sample space S of this experiment is S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
where (H, H) denotes that both the tosses result into
head and (T, i) denote the first toss result into a tail and
the number i appeared on the die for i = 1,2,3,4,5,6.
My book tells that all outcomes of this experiment are not equally likely and it assigned the probabilities to the 8 elementary
events
(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
as $\frac14 $, $\frac14 $, ${1 \over 12}$, ${1 \over 12}$,${1 \over 12}$, ${1 \over 12}$, ${1 \over 12}$, ${1 \over 12}$ respectively.
My question is why does it consider outcomes of this experiment to not be equally likely and if say they are not equally likely then why does it assign specifically these probabilities to the elementary events?
And why we don't solve the question as elementary events , which is :
Consider the experiment of throwing a die, if a simple multiple of 3 comes up, throw the die again and if any other number comes , toss a coin . Find the E/F where E : "the coin show a tail" and F : "at least one die shows a 3"
Best Answer
When you flip the first coin there are two equally probable results: $\rm H$ or $\rm T$. The probability for each is $1/2$.
Now if that result was a head (half of the total probability), you must flip the coin a second time, and again there are two equally probable results branching from that point. This give the probabilities of two of the outcomes $\rm (H,H), (H,T)$ as each being half of the half: $1/4$.
Now if the first toss were a tail, you would toss a die. This time their would be six outcomes branching off that initial result, all equally likely from that point. This give the probabilities of these remaining six outcomes $\rm (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)$ as each being $1/12$.
This is nothing more than the definition of conditional probability.
$$\mathsf P((X,Y){=}(x,y)) ~=~ \mathsf P(X{=}x)~\mathsf P(Y{=}y\mid X{=}x)\\\mathsf P((X,Y){=}({\rm T},6)) ~=~ \mathsf P(X{=}{\rm T})~\mathsf P(Y{=}6\mid X{=}{\rm T}) \\=~\tfrac 1 2\times \tfrac 16$$
And such.
PS: the probability that there die shows greater than 4 given that there is at least one tail is obviously: