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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 8outcomes 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 Tshould 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 Hand 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 Ymust 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.
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