Example 5.For the blood typing experiment of the above example. P(third type is A+)=P(third is first isn'tsecond isn't) .P(second isn't first isn't).P(first isn't) Example 6:A box contains w white balls and r red balls.Draw 3 balls without replacement.What is the probability of getting the sequence white,red,white? Solution: P(Win R2nW3)=P(Wi)P(R2|Wi)P(W3|R2nWi) (4）x(+)x() Example 7.Tom gets the bus to campus every day.The bus is on time with probability ,and late with probability0.4. The sample space can be written as n fbus journeysf. We can formulate events as follows:T=“on time”;L=“late” From the information given,the events have probabilities: PT=0.6;PL)=0.4 Question(a)Do the events T and L form a partition of the sample space? Explain why or why not. Solution. Yes. They cover all possible journeys (probabilities sum to 1),and there is no overlap in the events by definition.Example 5. For the blood typing experiment of the above example, 0.25 4 1 4 3 3 2 2 1 (sec ' | ' ) ( ' ) ( ) ( | ' sec ' )           P ond isn t first isn t P first isn t P third type is A P third is first isn t ond isn t Example 6: A box contains w white balls and r red balls. Draw 3 balls without replacement. What is the probability of getting the sequence white, red, white? Solution: Example 7. Tom gets the bus to campus every day. The bus is on time with probability 0.6, and late with probability 0.4. The sample space can be written as We can formulate events as follows: T = “on time”; L = “late”. From the information given, the events have probabilities: P(T) = 0.6 ; P(L) = 0.4 Question(a) Do the events T and L form a partition of the sample space? Explain why or why not. Solution. Yes. They cover all possible journeys (probabilities sum to 1), and there is no overlap in the events by definition