# How to find the probability of elementary events of a random experiment when they are not equally likely?

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"

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:

$$\mathsf P(Y\in\{5,6\}\mid X{=}{\rm T}\cup Y{=}{\rm T})=\dfrac{\mathsf P((X,Y)\in\{~({\rm T},5),({\rm T},6)~\}) }{\mathsf P((X,Y)\in\{~({\rm H},{\rm T}),({\rm T},1),({\rm T},2),({\rm T},3),({\rm T},4)({\rm T},5), ({\rm T},6)~\})}$$

Let's look at just one elementary event, to get the idea: What is the probability of $(T,1)$.

Well, first yo have to flip a tails -- that is probability $\frac12$ to happen. Then if you do, you have to roll a 1; that is probability $\frac16$ if you rolled a tails in the first place. So the probability of actually rolling a 1 is $$\frac12 \cdot \frac16 = \frac1{12}$$

One trap to avoid in any prob class is stubbornly assigning equal probabilities to all outcomes of an experiment without considering how those outcomes occur. As a more familiar example: Flip ten honest coins and count the number of heads. The probability of six heads and four tails is much greater than the probability of nine heads and one tail.

• why did you multiply $\frac 12$ and $\frac16$ to get probability of (T,1)
– Matt
May 13, 2016 at 22:36
• Because you can write $P(\{T\} \cap \{1\}) = P(\{T\})P(\{1\} \mid \{T\})$, by definition of conditional probability. May 13, 2016 at 22:44