How about tossing three heads successively ( HHH ) ? That would be 0.5 adam 0.5 ten 0.5, or ( 0.5 ) 3 How about tossing two heads and then one tail in that ordain ( HHT ) ? That would be 0.5 ten 0.5 ten 0.5, or ( 0.5 ) 3

How about tossing two tails and then one head in that order ( TTH ) ? That would again be ( 0.5 ) 3.

Doing more of these examples, you can convince yourself that the probability of getting a particular consequence tossing three coins in a row is ( 0.5 ) 3.

Let us now write out all the possible outcomes from tossing 3 coins in a course : { HHH, HHT, HTH, THH, TTH, THT, HTT, TTT }

In other words, there are 8 potential outcomes, and each result has a probability of ( 0.5 ) 3 If **X ** represents the number of heads that will result from the 3 tosses, **X ** can be 0, 1, 2, or 3 based on the above result ( that is, 0 heads can result, 1 lead can result, 2 promontory can result or 3 head can result ).

nowadays let ‘s consider these questions : What is the probability that X = 0 ? That would correspond to the consequence TTT, which would be ( 0.5 ) 3 What is the probability that X = 1 ? That would correspond to the outcomes HTT, THT, or TTH, depending on when the head comes. That would be ( 0.5 ) 3 + ( 0.5 ) 3 + ( 0.5 ) 3 or 3 ( 0.5 ) 3

A probability distribution is basically all the potential values of the random variable and their equate probabilities. sol for our casing, this would be the probability distribution :

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X = 0, ( 0.5 ) 3 X = 1, 3 ( 0.5 ) 3 X = 2, 3 ( 0.5 ) 3 X = 3, ( 0.5 ) 3

The union of all probabilities in a probability distribution should be 1, which is true in our casing.

Graphing this would be simple. Since this is a discrete routine, you should have to denote each value of x and its probability using vertical lines.

E ( X ) is basically the big value, or weighted average, of the probability distribution. To calculate the E ( X ), we need to sum x*p ( x ) for all values of x. indeed in this case, it would be 0* ( 0.5 ) 3 + 1*3 ( 0.5 ) 3 + 2*3 ( 0.5 ) 3 + 3* ( 0.5 ) 3, which is 1.5

V ( X ) is the discrepancy and is a measure of how far each random variable is from the beggarly. To calculate the discrepancy we need to average the square of the difference of each random variable and the mean. so, in other words, V ( X ) = E [ ( X-E ( X ) ) 2 ] sol, ( 0-1.5 ) 2 ( 0.5 ) 3 + ( 1-1.5 ) 2 3 ( 0.5 ) 3 + ( 2-1.5 ) 2 3 ( 0.5 ) 3 + ( 3 -1.5 ) 2 ( 0.5 ) 3 = 0.75 then we divide 5 by the number of trials, which in this subject was 3 ( since we tossed the coin 3 times ). so 5/3 is the variation

note : this is an exemplar of the binomial distribution ! You can read about it foster on-line. Knowing that it is a binomial distribution can provide many useful shortcuts, like E ( X ) = neptunium, where newton = 3 and p = 0.5, or V ( X ) = neptunium ( 1-p )