tgt

## Thursday, 7 August 2014

### chapter-3 - Example of Tossing a Coin

Scenario -2:
You toss a fair coin thrice. Since there are two outcomes at each toss, namely “Heads” and “Tails” the total number of possible outcomes is
 $2 \times 2 \times 2 = 8$
Let us list down all the $8$ outcomes for clarity’s sake; $H$ represents “Heads ” and $T$ represents “Tails”:
 $HHH,\, HHT, \, HTH, \,THH$ $HTT, \, THT, \, TTH, \, TTT$
Since the coin is fair, we have sufficient reason to believe that each of these $8$outcomes is equally likely, i.e. equi-probable. Also, since these are the only outcomes possible and one of them must occur, we can say that the probability of any outcome is $\dfrac{1}{8}$. For example,
 $P\left( {HHT} \right) = P\left( {TTH} \right) = \dfrac{1}{8}$
Now, let us pose the following question: in the sequence of these three throws, what is the probability of $H$ occuring before $T$?
This is easy; we count all cases satisfying the stated constraint:
 $HHH, HHT, HTH, HTT$
Since there are $4$ “favorable” cases to our constraint out of the total $8$ cases possible, it should be correct to say that the possibility of $H$ occuring before $T$should be $\dfrac{4}{8} = \dfrac{1}{2}$. Similarly, the possibility of $T$ occuring before $H$ should also be $\dfrac{4}{8} = \dfrac{1}{2}$.
Now let us try to answer the same question from a slightly different perspective. Let us define the events $E$ and $F$ as follows for this sequence of three tosses:
 $E$ : $H$ occurs before $T$ $F$ : $T$ occurs before $H$
The task is to find $P(E)$ and $P(F)$, which we earlier did by explicitly counting the favorable cases. Here we’ll try to calculate $P(E)$ and $P(F)$ differently. Observe that one of the events $E$ or $F$ must occur; there is no other possibility. If we denote the outcomes of our “experiment” (involving the sequence of three tosses), by a rectangle, we can divide the rectangle into exactly two halves, one representing the event $E$ and one the event $F$.
Now, it should be obvious that
 $P\left( E \right) + P\left( F \right) = 1$ $\ldots(3)$
Also, since the coin is fair, $E$ and $F$ must be equi-probable. This is because $H$and $T$ are equi-probable, i.e, equivalent outcomes in terms of probability. Thus, $H$ occuring before $T$ must be as likely as $T$ occuring before $H$. Hence, we must have
 $P\left( E \right) = P\left( F \right)$ $\ldots(4)$
From $(3)$ and $(4)$, we must have
 $P\left( E \right) = P\left( F \right) = \dfrac{1}{2}$
Let us now pose a further question for this sequence of three tosses. Let the events $X$ and $Y$ be defined as follows:
 $X$ : There are more $H$ than $T$ $Y$ : There are more $T$ than $H$
Our task is to calculate $P(X)$ and $P(Y)$. These should be immediately apparent. Since there are an odd number of tosses, one of the events $X$ or $Y$must occur. Also, $X$ and $Y$ have to be equi-probable.
Thus,
 $P\left( X \right) + P\left( Y \right) = 1$ and $P\left( X \right) = P\left( Y \right)$ $\Rightarrow \,\,\,\, P\left( X \right) = P\left( Y \right) = \dfrac{1}{2}$
Your are urged to arrive at the same result by the explicit counting of out comes.