Example:
Two-dice totals. There are 36 possible simple events when two dice are thrown
but a gambler may only want to know the total.
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| Click on the variable for an example | |
| 2 | 0.028 |
| 3 | 0.056 |
| 4 | 0.083 |
| 5 | 0.111 |
| 6 | 0.139 |
| 7 | 0.167 |
| 8 | 0.139 |
| 9 | 0.111 |
| 10 | 0.083 |
| 11 | 0.056 |
| 12 | 0.028 |
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Following the above values for X, we would need Prob(0) , Prob(1) , Prob(2),
... to be assigned proper values.
Requirements for a probability distribution:
1. Every probability that is given , say Prob(X), must
be between 0% and 100%.
Again, 0% < Prob(X) < 100% or 0
< P(X) < 1.00
2. The sum of all these probabilities must be 1.00 or
100%.
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Typically we use X as the random variable.
A discrete random variable has either a finite number of values or a countable number of values.
For example, we may use
X = 0 , 1 , 2 , 3 , 4 , ....
for the discrete random variable.
A probability distribution gives the probability for each value of the random variable.
Question:Part 1)
Yes, there are 11 discrete values of the random variable X. (from 2 to 12) and
the probabilities are all between 0 and 1.00 .
Part 2)
The sum of all the probabilities is 1.00. (Actually add up the numbers and you
get 1.001--This is a rounding error. More decimal places could have been used.)

Example:
We are given random variable X values of 0,1,2,3,4,5,6,7.
Probabilities are given :
P(0)= 0.09 P(1)= 0.16 P(2)= 0.21 P(3)=0.19 P(4)=0.14
P(5)= 0.08 P(6) = 0.08 P(7)=0.05
Question:
Make
a summary table of X and P(X) values. Make a histogram.
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Probability distributions are a basis for much of Probability and Statistics.
For example, if you could see a probability distribution of Temperature for
the month of January then you could plan which temperatures you are most likely
to encounter.
Some probability distributions have formulas:
Example:
Prob(X) = 2-x
X =1,2,3,4,5,6,.....
Question:
Prove
this is a probability distribution.
Make a histogram.
Easiest to, again, make a Summary Table
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| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| Toss 1 | ||||||||||||||||
| Toss 2 | ||||||||||||||||
| Toss 3 | ||||||||||||||||
| Toss 4 | ![]() |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| Toss 1 | ||||||||||||||||
| Toss 2 | ||||||||||||||||
| Toss 3 | ![]() |
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| Toss 4 | ![]() |
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| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| Toss 1 | ||||||||||||||||
| Toss 2 | ||||||||||||||||
| Toss 3 | ![]() |
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| Toss 4 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| Toss 1 | ||||||||||||||||
| Toss 2 | ||||||||||||||||
| Toss 3 | ||||||||||||||||
| Toss 4 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | |
| Toss 1 | ||||||||||||||||
| Toss 2 | ![]() |
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| Toss 3 | ||||||||||||||||
| Toss 4 | ![]() |
Question :
What is the probability of
4 "Heads" out of four tosses ?
How
about 3 "Heads" out of 4?
Tossing
2 "Heads" out of4 ?
Tossing
1 "Heads"out of 4 ?
Tossing
0 "Heads" out of4 ?We
need a random variable X which is properly defined:
X = number of "Heads" out of 4 .
Clearly values of X are 0 , 1 , 2 ,3 , 4
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0
1/16
6.25%
1
4/16
25.0%
2
6/16
37.5%
3
4/16
25.0%
4
1/16
6.25%
Part 1)
Yes all 5 probabilities are between 0 and 1.00 since they range from a low
probability of 0.0625 to a highest probability of 0.3750 .
Part 2)
Add up the five probabilities to get 1.0000.
Conclusion:
This is a clearly defined probability distribution.