implies that p(b |a) = p(b). Likewise, the probability of a knowing b must also be unchanged, or p(a |b) = p(a). Replacing these two results in Eq. (1.16) yields a fundamental property for the joint probability of independent/uncorrelated events: p(a,b)= pla)p(b). (1.17) In the opposite case to correlated or dependent events, we have p(a, b) # p(a)p(b), which means that either p(a,b) > p(a)p(b) or p(a, b) < p(a)p(b), representing two possibilities (the second case is being referred to as anti-correlation). Let us play with conditional probabilities through an illustrative example. Party meetings Assume that the probabilities of meeting two old friends, Alice and Bob, in a party are p(Alice) = 0.6 and p(Bob) = 0.4, respectively. The probability of meeting Bob if we know that Alice is there is |Alice) = 0.5. Then what are the probabilities of: (a) Meeting at least one of them? (b) Meeting both Alice and Bob? (c) Meeting Alice if we know that Bob is there? Answer: For question (a), let's apply the theorem in Eq. (1.13) combined with Bayes's theorem in Eq. (1.16) to get: p(Alice + Bob) = p(Alice) + p(Bob) - p(Alice, Bob) = p(Alice) + p(Bob) - p(Bob |Alice)p(Alice). = 0.6 + - 0.5*0.6 = 0.7 For question (b), we directly calculate the joint probabilities as follows: p(Alice, Bob) = |Alice)p(Alice) = 0.5*0.6 = 0.3. For question (c), we apply Bayes's theorem get: We see from the results that the most likely outcome is to meet at least one of them, (p(Alice + Bob) = 0.7), meaning either Alice or Bob or both. The less likely outcome is to meet both (p(Alice, Bob) = 0.3). But if Alice is first seen there, it is much more likely to see Bob (p(Bob|Alice) = 0.7 > p(Bob) = 0.4) and if Bob is first seen there, it is a bit more likely to see Alice (p(Alice |Bob) = 0.75 > p(Alice) = 0.6). These last two facts Probability basics illustrate the correlation between the respective presences of Alice and Bob. If their presence were not correlated, we would have p(Alice, Bob) = p(Alice) X p(Bob) = 0.6*0.4 = 0.24, which is a lower probabi