2. Probability theory The department of math of central south university Probability and Statistics Course group

3. The first chapter Exercise Example 1 Assuming there are three boxes of similar models of parts, which were equipped with 50, 30 and 40, and first-class goods were 20, 12 and 24. Now choose from a case of random samples of all has a spare parts (for the first time to take back the parts do not) try to seek access to spare parts is the probability of first-class goods and out calculating both the probability of first-class goods.

4. Solution: set respectively in the first and second election of the first-class products with pieces of the box for 20,12,24, respectively for the first time, for the second out of first-class products, according to Italian title,

5. Example 2 A, B, C at the same time of three aircraft fire, hitting three of the probability of 0.4,0.5,0.7. Planes were shot down and one hit a probability of 0.2 and was hit by two shot down by a probability of 0.6 If all three hit, the aircraft must have been shot down, and the probability of the aircraft was shot down. Solution: Let B = ( plane was shot down), Ai = (aircraft was hit i people), i = 1,2,3

6. By the whole probability formula P (B) = P (A1) P (B | A1) + P (A2) P (B | A2) + P (A3) P (B | A3) = 0.36 × 0.2 +0.41 × 0.6 +0.14 × 1 = 0.458 That the aircraft was shot down by a probability of 0.458.

7. Example 3 From the assumption that there are three regions of the 10, 15 and 25 candidates of the nomination form, the application forms which girls were 3, 7 and 5. Are randomly out of a region of the entry form, and from one after another at random Two out. (1) to seek out a female form of probability; (2) known after the draw is a table of boys,to seek out a female form of probability.

8. solution: introduction of events ： { application forms is the first region j} ； ={ I is the first time be able to get into the boys table}. by the conditions of knowledge: by the probability of the whole formula, we have

9. Easy to see by the probability of the whole formula:

10. by the definition of conditional probability

11. Example 4 set the probability of the same events and independent of each other, seek Solution set ． and the formula by adding event independent,we have

12. Example 5 Prove events that the full independence and a necessary condition is that they are pairwise independent and any of the events and the remaining two events of the independent pay. Proof: (1) need Events for A, B, C independent, any two of them independent of each other; In addition, by P (A [BC]) P (ABC) = P (A) P (B) P (C) = (A) P (BC), A visible with the independent BC. The same evidence, B and AC independent, C and AB independence.

13. (2) Adequacy events pairwise independent and any of the events and the rest of the second incident of cross-independent. Pairwise independent by the independent, to know So events independent ． A result of BC and independence, can be seen

14. A short break to continue