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P(E) = n(E) / n(S) = 3 / 36 = 1 / 12 c) All possible outcomes, E = S, give a sum less than 13, hence. P(E) = n(E) / n(S) = 1 / 4 Question 3: Which of these numbers cannot be a probability? a) -0.00001 b) 0.5 c) 1.001 d) 0 e) 1 f) 20% Solution to Question 3: A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1. Solution to Question 1: Let us first write the sample space S of the experiment. An examination of the sample space shows that there are 4 "Queens" so that n(E) = 4 and n(S) = 52. Hence the probability of event E occurring is given by P(E) = 1 / 52 Question 7: A card is drawn at random from a deck of cards.

b) Two coins are tossed, find the probability that one head only is obtained. If a marble is drawn from the jar at random, what is the probability that this marble is white? Solution to Question 8: We first construct a table of frequencies that gives the marbles color distributions as follows color frequency red 3 green 7 white 10 We now use the empirical formula of the probability Frequency for white color P(E)= Total frequencies in the above table = 10 / 20 = 1 / 2 Question 9: The blood groups of 200 people is distributed as follows: 50 have type A blood, 65 have B blood type, 70 have O blood type and 15 have type AB blood. Question 1: A die is rolled, find the probability that an even number is obtained. Event E may be described as follows E={(1,H),(3,H),(5,H)} The probability P(E) is given by P(E) = n(E) / n(S) = 3 / 12 = 1 / 4 Question 6: A card is drawn at random from a deck of cards. In what follows, S is the sample space of the experiment in question and E is the event of interest. E = {2,4,6} We now use the formula of the classical probability. S = {(H,T),(H,H),(T,H),(T,T)} Let E be the event "two heads are obtained". Note: Each coin has two possible outcomes H (heads) and T (Tails). 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".

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