Introduction for probability:
Probability tables is a quite similar to chance In general, to obtain the probability of an event, we find the ratio of the number of outcomes favourable to the event, to the total number of equally likely outcomes.
Experiment: An experiment is defined as a process for which its result is well-defined.
Random experiment: An experiment whose all possible outcomes are known, but it is not possible to predict the outcome.
Distribution function: (Cumulative Distribution function)
The distribution function for random variable X is defined as
F(x) = P(X ≤ x) =`sum_(xi<x)^`
Probability Mass Function:
The discrete probability tables function p(x) which satisfy following propertise:
(1) The probability tables that X can take a specific value x is p(x)
ie., P(X = x) = p(x) = px.
(2) p(x) is non – negative for all real x.
(3) The sum of p(x) over all possible values of X is one. That is
Σpi = 1 where j represents all possible values that X can have and pi is the probability at X = xi
If a1, a2, . . . am, a, b1, b2, . . bn, b be the values of the discrete random variable X in ascending order then
(i) P(X ≥ a) = 1 − P(X < a)
(ii) P(X ≤ a) = 1 − P(X > a)
(iii) P(a ≤ X ≤ b) = P(X = a) + P(X = b1) + P(X = b2) + . . .. . . + P(X = bn) + P(X = b).
Example problem in probability tables:
A random variable X has the following probability tables mass function
(1) Find k.
(2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X ≤ 6)
(1) Since P(X = x) is a probability tables mass function P(X = x) = 1`sum_(x=0)^6`
ie.,P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.
⇒ k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ 49 k = 1 ⇒ k =1/49
(2) P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3)
P(X ≥ 5) = P(X = 5) + P(X = 6) =(11/49)+(13/49)=24/49
P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6) =(9/49)+(11/49)+(13/49)=33/49
A random variable X has the following probability table mass function
(1)What is the smallest value of x for which P (X ≤ x) >1/2.