S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}, where B stands for a defective or bad pen and G for a non – defective or good pen. Let the probabilities assigned to the outcomes be as follows Sample point: BBB BBG BGB GBB BGG GBG GGB GGG Probability: Let event A: there is exactly one defective pen and event B: there are atleast two defective pens. Hence A = {BGG, GBG, GGB} and B = {BBG, BGB, GBB, BBB} Now P(A) = P( ω ), ω A ∑ ∀ ∈ = P(BGG) + P(GBG) + P(GGB) = and P(B) = P( ω ), ω B ∑ ∀ ∈ = P(BBG) + P(BGB) + P(GBB) + P(BBB) = Let us consider another experiment of ‘tossing a coin “twice” The sample space of this experiment is S = {HH, HT, TH, TT} Let the following probabilities be assigned to the outcomes PROBABILITY P(HH) = , P(HT) = , P(TH) = , P(TT) = Clearly this assignment satisfies the conditions of axiomatic approach. Now, let us find the probability of the event E: ‘Both the tosses yield the same result’.
Here E = {HH, TT} Now P(E) = Σ P( w i ), for all w i ∈ E = P(HH) + P(TT) = For the event F: ‘exactly two heads’, we have F = {HH} and P(F) = P(HH) = . . Probabilities of equally likely outcomes Let a sample space of an experiment be S = { ω , ω ,..., ω n }. Let all the outcomes are equally likely to occur, i.e., the chance of occurrence of each simple event must be same.
i.e. P( ω i ) = p , for all ω i ∈ S where ≤ p ≤ Since P( ω ) i =