IfE 2 is the event that an 11 is rolled, thenP(E 2 ) =P((5,6)) +P((6,5)) = 2/36 = 1/ 18IfE 3 is the event doubles are rolled, thenP(E 3 ) =P((1,1)) +P((2,2)) +P((3,3)) +P((4,4)) +P((5,5)) +P((6,6)) = 6/36 = 1/ 6IfE 4 is the event that a 10 is rolled, thenP(E 4 ) =P((4,6)) +P((5,5)) +P((6,4)) = 3/36 = 1/ 12If E 5 is the event a 10 is rolled or a double is rolled, the E 5 =E 4 ∪E 3. Notethat the event(5,5)is common to both eventsE 4 andE 3 withE 4 ∩E 3 = (5,5)and
P(E 4 ∩E 3 ) =P((5,5)) = 1/ 36. Hence,
P(E 5 ) =P(E 4 ∪E 3 ) =P(E 4 ) +P(E 3 )−P(E 4 ∩E 3 ) = 3/36 + 6/ 36 − 1 /36 = 8/36 = 2/ 9If E 6 is the event that a 10 is rolled or a 7 is rolled, thenE 6 =E 4 ∪E 1. HereE 4 ∩E 1 =∅and so these events are mutually exclusive. Consequently,
P(E 6 ) =P(E 4 ∪E 1 ) =P(E 4 ) +P(E 1 ) = 3/36 + 6/36 = 9/36 = 1/ 4Note that there are many situations where the sample space is not finite. Insuch cases the probabilities assigned to the events inSarebased upon employment
of relative frequencies observed from taking a large numberof trials from the pop-
ulation being studied. For a large number of trials the relative frequency of an
event happening is approximately the same as the probability of the event happening.
Observation of this fact should be made by examining the earlier development of
relative frequency tables associated with our analysis of data collected. Also note
that one can think of equation (11.20) is a special case of themore general equation
(11.22), the special case occurring whenE 1 andE 2 are mutually exclusive events.
If pis a probability of success assign to a single trial of an event, then theexpected numberof successes inn-trials is the productnp.