lity than found here (p (Alice, Bob) = 0.3). This shows that the known presence of either one increases the chances of seeing the other, a fact that is usually verified in parties or social life! Incidentally, the probability that neither Alice nor Bob be seen at the party is p(Alice + Bob) = 1 - p(Alice + Bob) = 1 - 0.7 = 0.3. Thus, there exist equal chances of seeing them both or of seeing neither of them, but this is only coincidental to this example. I shall conclude this section and chapter by considering the probabilities associated with a succession of independent or uncorrelated events. As we have seen, if X1, X2 are independent events with probabilities p(x1), p(x2), their joint probability is simply given by the product p(x1, X2) = p(x1) X p(x2). Thus for three independent events, we have p(x1, X2, X3) = p(x1, x2) X p(x3) = p(x1)p(x2)p(x3), and SO on. Consider a few examples to illustrate the point. Tossing the coin What is the probability that a flipped coin lands three times on the same side? Answer: The probability of getting either three tails or three heads is q = (1/2)3 =0.125. But there is a trick: since either succession of events is possible (i.e., getting three heads or three tails), the total probability is actually 0.125 + 0.125 = 0.250 = 1/4. Double six Rolling two dice, what is the number of trials required to obtain a double six with a chance of at least 50%? Answer: The probability of getting a six from a single die is p(6) = 1/6. Since the two dice are independent (their outcome is uncorrelated), the joint probability for a double six is p(6,6) = p(6) X p(6)(1/6)2 = 1/36. The probability of not getting a double six is therefore q = 1 - p(6, 6) = 1 - 1/36 = 35/36 = 0.97222 The probability of not getting a double six after N trials is qN. We must then find the number N, such that qN < 0.5, meaning that the complementary event (not getting any double six) has the proba- bility 1 - qN > 0.5, or greater than 50%. Solving this equation for N is straightforward using the successive steps: N lo