Some Operating cases on Probabilities

                   Some Operating cases on Probabilities

 * It is suggested that operations on probabilities are a bit different from numeric operations because the range of occurrence of such probability values generally lies between the range of 0 & 1

 

* One must rely on some set or rules in order for the operation to make sense to the user who is conducting the experiment on probabilities. For example , if someone is conducting an experiment of tossing a coin then he/she must strictly define the rules according to which the game of tossing a coin would be played out . The instructor would declare which outcomes should be taken as valid outcomes and which should not be taken in as valid outcomes , rather must be negated the moment the norms of the game are violated .

 * For example , suppose say a case happens over where a coin does not fall over any of the sides rather falls over the floor standing erect , then the outcome is neither a heads and nor a tails , and neither a 50-50 heads-tails can be taken as consideration for the throw of the dice . Rather what would happen in such a circumstance is that the throw of the dice for this case would be nullified , the entire event of throw of such a dice would be struck off from the probable set of outputs that should happen as a result of the throw of the dice . Thats why one should also keep adhering to the rules of the experiment before conducting such an experiment which would require to know what set of events should be taken in as considerable outcomes and which should not be considered .

 

* Again another property of Probabilities that one needs to be aware is summations between probabilities which states that summations of probabilities is possible only when all the constituting events of the sample space are mutually exclusive to each other . For example lets consider an experiment of rolling a dice over a game of ludo , in this all the possible events that could turn up as a result of throw of the dice are 1 , 2 , 3 , 4 , 5 , 6 . The probability of occurrence of each of the events is 1/6 or 1 by 6 . And here , each of the events within the given sample space are disjoint and mutually exclusive to each other which makes the individual events probability of occurrence as equal to each of the given event divided by the total number of events over the entire sample space . And in case one would like to know the probability of occurrence of all the events together in unison , then one may have to add up the probabilities of each of the individual events as a summation of each of the individual events .. which would yield an output of 1 . So in retrospect, all individual elements of an experiment of probability are disjoint and mutually exclusive and in unison lead to a summed up value of 1 .

 

* We can take another simple example to demonstrate to demonstrate the case of understanding of probability calculation ; in this case one can consider for example the case of picking a spade or a diamond from a set of cards can be calculated in the following manner . Total number of cards in the entire deck = 52 . Number of cards in the house of clubs = 13 , number of cards in the house of clubs = 13 , number of cards in the house of hearts = 13 , number of cards in the house of diamonds = 13 . If a person takes out a card from the house of diamond then the probability if picking up one of the cards is 13/52 ; the same goes for the case of picking up a random card from a house of clubs is 13/52 . So , total probability of finding a card from both the houses is 26/52 which is equals to 0.5

 

* One can take the help of subtraction operation to determine the probability of some events where probability of an event is different from the probability of an event that one would want to compare . For instance , if someone wants to determine the probability of drawing a card that does not belong to some house of card for example , say I want to draw a card which is not a diamond from the overall deck of cards , then one will approach the problem in the given manner . He will first find out the overall probability of finding any card and then he will subtract the chance of occurrence of a particular card from the total , 1 - 0.25 which happens to be as 0.75. One could get a complement of the occurrence of the card in this manner , which could be used for finding the probability of not occurrence of a particular event .

 

* Multiplication of a set of events can be helpful for finding the intersection of a set of independent events . Independent Events are those which do not influence each other . For instance , if one is playing a game of dice and one would like to throw two dices together , then the probability of getting two sixes is 1/ 36 . This can be obtained by multiplication of dices over both the cards , where first the probability of obtaining a 6 is found out to be as 1/6 and then the subsequent independent event would also produce an probability of obtaining another 6 is found out to be as 1/6 , here both the values are multiplied with each other and found that product of both the probabilities of independent events would yield a value output as 1/36 or 0.28 .

 

* Using the concepts of summation , difference and multiplication , one can obtain the probability of most of the calculations which deal with events . For instance , if one would want to compare the probability of getting atleast a six from two throws of dice which is a summation of mutually exclusive events . Probability of obtaining two sixes of dice , p = 1/6* 1/6 = 1/36

 

* In a similar manner if one would like to calculate the probability of having a six on the first dice and then something other than a six on the second throw of the dice is p = (1/6)*(1- 1/6) = 5/36 ,

 

* Probability of getting a six from two thrown dice is p = 1/6* 1/6 +2*1/6*(1- 1/6) = 11/36