Computers and Technology
Submitted By Shubham2609
ANALYSIS OF DATA
• An experiment is an occurrence whose result, or outcome,
• The set of all possible outcomes is called the sample space for the experiment.
• Given a sample space S, an event E is a subset of S.
• The outcomes in E are called the favourable outcomes.
• We say that E occurs in a particular experiment if the
outcome of that experiment is one of the elements of E.
• When a fair die is rolled, probability of each outcome is 1/6.
• So probability of E: Even numbers is P(E) = 3/6 = 0.5.
• But a die does not need to be fair, and outcomes are not
always equally likely!
• For example, suppose we have a loaded die with probabilities of 1, 2, 3, 4, 5 and 6 resp. 0.25, 0.15, 0.15, 0.15,
0.15 and 0.15.
• Then P(E) = 0.45.
For any event A:
i) 0 ≤ P(A) ≤ 1. ii) P(Φ) = 0, where Φ is the null event (no outcomes). iii) P(S) = 1, where S is the sample space (all outcomes). iv) If A and B are two events with no common outcome, (i.e.
“mutually exclusive”) then
P(A B) P(A) P(B)
v) If A and B are two mutually exclusive events such that
P(A B) 1 then A and B are called complements of each other. We denote B as A or AC.
P(A B) P(A) P( B) P(A B)
• Consider the fair die being rolled twice
• P(sum is even) =
• P(product is odd) =
• P(sum is even, product is odd) =
• Re-compute the probabilities for the loaded die.
Work out the probabilities
i) P(sum is even given the product is odd) ii) P(product is odd given sum is even) for the fair die as well as the loaded die.
How to proceed?
• Let A, B be two events such that P(A) > 0. Then
P ( A B)
P( B | A) …...