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Enumerated Type TContactViewEvent::TEventType This is used to specify the type of event to a view observer. Some of these event types cannot apply to a local view. Some of the event types indicate that further information is available in the relevant TContactViewEvent object. EUnavailable The observed view s state has changed from EReady to either ENotReady or EInitializing, so is not available for use. EReady The observed view s state has changed from ENotReady or EInitializing to EReady so is available for use. ESortOrderChanged The observed view s sort order has changed, so observer views need to update themselves. ESortError An error occurred when sorting the observed view or when appending an observer to its observer array. The error code is provided in iInt. EServerError An error occurred in the contacts server. The error code is provided in iInt. EIndexingError An error occurred when setting the range for a CContactSubView. The error code is provided in iInt. EItemAdded An item has been added to the observed view. The identi er of the added item is provided in iContactId, and the index into the underlying view of the added item is provided in iInt. EItemRemoved An item has been removed from the observed view. The identi er of the removed item is provided in iContactId, and the index into the underlying view of the item is provided in iInt. EGroupChanged A change has occurred in a contact group, for instance a contact item has been moved into or out of the group. The identi er of the group affected is provided in iContactId.

Note that the probabilities in the table add up to 1 as they should Now consider the random variable X + Y This random variable can take on the values 2, 3, 4, or 5 There is only one way for X + Y to be 2, namely, each of the variables must be 1 So the probability that X + Y = 2 is 1/12 and we write P(X + Y = 2) = 1/12 There are two mutually exclusive ways for X + Y to be 3, namely, X=1 and Y =2 or X=2 and Y = 1 So P(X + Y = 3) = P(X = 1 and Y = 2) + P(X = 2 and Y = 1) = 1/3 + 1/12 = 5/12 It is easy to check that there are two mutually exclusive ways for X + Y to be 4 and this has probability 5/12 Finally, the probability that X + Y = 5 is 1/12 These probabilities add up to 1 as they should This means that the random variable X + Y has the following probability distribution function: 1/12 if x + y = 2 5/12 if x + y = 3 f (x + y) = 5/12 if x + y = 4 1/12 if x + y = 5 where x and y denote values of the random variables X and Y This random variable then has a mean value We nd that 1 5 5 1 7 E(X + Y ) = 2 +3 +4 +5 = 12 12 12 12 2 How does this value relate to the expected values of the variables X and Y taken separately First, we must nd the probability distribution functions of the variables alone What, for example, is the probability that X = 1 We know that P(X = 1 and Y = 1) = 1/12 and P(X = 1 and Y = 2) = 1/3 These events are mutually exclusive and are the only events for which X = 1 So P(X = 1) = 1/12 + 1/3 = 5/12 Notice that this is the sum of the probabilities in the column in Table 81 for which X = 1 In a similar way, P(X = 2) = P(X = 2 and Y = 1) + P(X = 2 and Y = 2) = 1/12 + 1/12 = 1/6, which is the sum of the probabilities in the column of Table 81 for which X = 2 Finally, summing the probabilities in Table 81 for which X = 3 is 1/3 + 1/12 = 5/12 So the probability distribution for the random variable X alone can be found by adding up the entries in the columns; the probability distribution for the random variable Y alone can be found by adding up the probabilities in the rows of Table 81.

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