Question 1:A die is rolled, find the probability that an even number is obtained.SolutionLet us first write thesample spaceS of the experiment.S = {1,2,3,4,5,6}Let E be theevent"an even number is obtained" and write it down.E = {2,4,6}We now use the formula of theclassicalprobability.P(E) = n(E) / n(S) = 3 / 6 = 1 / 2

Question 2:Two coins are tossed, find the probability that two heads are obtained.Note:Each coin has two possible outcomes H (heads) and T (Tails).SolutionThe sample space S is given by.S = {(H,T),(H,H),(T,H),(T,T)}Let E be the event "two heads are obtained".E = {(H,H)}We use the formula of the classical probability.P(E) = n(E) / n(S) = 1 / 4

Question 3:Which of these numbers cannot be a probability?a) -0.00001b) 0.5c) 1.001d) 0e) 1f) 20%A probability is always greater than or equal to 0 and less than or equal to 1, hence onlya)andc)above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1

Question 4:Two dice are rolled, find the probability that the sum isa) equal to 1b) equal to 4c) less than 13Solutiona) The sample space S of two dice is shown below.S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, henceP(E) = n(E) / n(S) = 0 / 36 = 0b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.P(E) = n(E) / n(S) = 3 / 36 = 1 / 12c) All possible outcomes, E = S, give a sum less than 13, hence.P(E) = n(E) / n(S) = 36 / 36 = 1

Question 5:A die is rolled and a coin is tossed, find the probability that the die shows an odd number and the coin shows a head.SolutionThe sample space S of the experiment described in question 5 is as followsS = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}Let E be the event "the die shows an odd number and the coin shows a head". Event E may be described as followsE={(1,H),(3,H),(5,H)}The probability P(E) is given byP(E) = n(E) / n(S) = 3 / 12 = 1 / 4