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1.4 The Axioms of Probability Frequency definition of probability Equally Likely Outcomes Definition 1.9 A Probability Measure on a sample space S is a function P which assigns a number P(A) to every event A in S in such a way that the following three axioms are satisfied: Axiom 1. P(A)≥0 for every event A. Axiom 2. P(S) =1 Axiom 3. Countable Additivity i.e. if A1,A2,…is an infinite sequence of mutually exclusive (disjoint) events then P( i 1 [or P ( A1 A2 Ai ) P ( Ai ) i 1 ) P ( A1 ) P ( A2 ) ] Example 1.10 A coin is tossed twice. What is the probability that at least one head occurs? Solution: Example 1.11 A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a umber less than 4 occurs on a single toss of the die, find P(E). Solution: Example 1.12 In Example 1.11, let A be the event that an even number turns up and let B be the event that a number divisible by 3 occurs. Find P ( A B) and P ( A B) Solution: Properties of Probability 1) P(Ø)=0. 2) Finite Additivity i.e. if A1 , A2 ,, An are mutually exclusive (disjoint) events n n then P Ai P ( Ai ) i 1 i 1 3) P ( AC ) 1 P ( A). 4) If B A then P ( A B ) P ( A) P ( B ) and P ( B ) P ( A) 5) P(A)≤1 for all events A. 6) If A and B are arbitrary events then P ( A B ) P ( A) P ( B ) P ( A B ) and hence P ( A B ) P ( A) P ( B ). Equally Likely Outcomes Theorem Let S be the sample space of an experiment. If S has n elements which are all equally likely, then for any event A S | A| P ( A) n where |A| denotes the number of elements in A. Model One: Drawing ball (1) Drawing without replacement Problem 1 There are 4 white balls and 2 black balls in a bag. Draw two balls consecutively out of the bag. Now find the probability when the two balls are white. Solution: (2) Drawing with replacement Problem 2 There are 4 red balls and 6 black balls in a bag. Draw consecutively 3 balls with replacement. Find Find the probability when the first two balls drawn are black and the third ball drawn is red. Solution: Class Exercise 1o Phone number Given that a phone number is made up of 7 digits, while the first digit can not be 0, find the probability that the number 0 appears exactly 3 times. ( Solution : p C 91 C 63 93 9 106 ) 2o Dice 3 die are cast, find the probability that the numbers you get after you throw the die add up to 4. ( Solution : p 3 63 ) Model Two: Ball placed in cylinder (1) No restriction on the capacity of cylinder Problem 1 Four balls are to be placed inside three cylinders. Assuming that there is no restriction on the number of balls each cylinder can contain, find the probability that cylinders 1 and 2 each contains 2 balls. Solution: (2) Each cylinder can only have 1 ball Problem 2 Four balls are to be placed into ten cylinders. Knowing that each cylinder can only contain one ball, find the probability that for the first 4 cylinders, each one contains a ball. Solution: Class Exercise 1o Room arrangement There are three people Mr. A, B and C who are going to be arranged into three rooms. Find the probability that each room has just one people in it. 2 ( Solution : ) 9 2o Birthday A class has a total of 20 students, who are all born in the same year. Find the probability that out of the 20 students, 10 were born on January 1st and 10 were born on December 31st. 20 10 ( Solution : p 36520 ) 10 10 Solution: The sample space for this experiment is S { HH , HT ,TH ,TT } Therefore, we assign a probability of w to each sample point. Then 4w=1,or w=1/4. If A represents the event of at least one head occuring then A { HH , HT ,TH } Thus 你认为哪个 结果出现的 可能性大? 1 1 1 3 P ( A) . 4 4 4 4 Solution: The sample space is S={1,2,3,4,5,6}. We assign a probability of w to each odd number and a probability of 2w to each even number. Since the sum of the probabilities must be 1, we have 9w=1 or w=1/9. Hence probabilities of 1/9 and 2/9 are assigned to each odd and even number, respectively. Therefore, and E {1,2,3} P ( E ) 1 / 9 2 / 9 1 / 9 4 / 9. Solution: For the events A={2,4,6} and B={3,6}, we have A B {2,3,4,6} and A B {6} By assigning a probability of 1/9 to each odd number and 2/9 to each even number, we have P( A B) 2 / 9 and P( A B) 2 / 9 1 / 9 2 / 9 2 / 9 7 / 9 Solution: Suppose A {both balls are white}, The number of all events in the sample space is C 62 The number of basic events in A is C 42 ,So P ( A) C 42 / C 62 Solution: Suppose A { Black on the first 2 draws, red on the third draw } Red on the third draw 4 First draw second draw Third draw 6种 Black on on the the second first draw Black draw 6 10 Total number of basic/simple events 10 10 10 103 , A contains 6 6 4 24 events, So 6 6 4 0.144 P ( A) 3 10 3 3 3 3 Solution: According to this type of placement, there are 3 3 3 3 34 ways to place the 4 balls into 3 cylinders. C42 2 C22 2 Therefore, the probability that cylinders 1 and 2 will each have 2 balls is: 2 2 2 4 p C4 C2 3 . 27 Problem 2: Four balls are to be placed into ten cylinders. Knowing that each cylinder can only contain one ball, find the probability that for the first 4 cylinders, each one contains a ball. 4 3 2 1 1 p 44 . Solution: p 4 p10 10 9 8 7 210