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Mathematical touch ball problem
( 1) ? , reference analysis; (2) Reference analysis

Test and analysis: (1) Four balls in the bag, 1 marked with face value of 50 yuan, and the other three balls are 10 yuan, requiring each customer.

A bag with four balls marked with face value randomly draws two balls at a time, and the sum of the face values marked on the balls is the customer's reward amount. Just divide the number of events that get 60 yuan by the total number of events. There are two ways for customers to get rewards: 20 yuan and 60 yuan. Calculate their probabilities respectively, and then draw a conclusion by using the formula of mathematical expectation.

(2) According to the budget of the mall, each customer will be rewarded with 60 yuan on average. According to the meaning of the question, there are two kinds of rewards, and there is only one scheme that meets the meaning of the question. Then calculate the probability, mathematical expectation and variance corresponding to the two schemes respectively, and then draw a conclusion.

Problem analysis: (1) Let the customer's reward be X. ① According to the meaning of the problem, the probability that the customer's reward amount is 60 yuan is.

② According to the meaning of the question, all possible values of X are 20,60 ... That is to say, the distribution list of X is

X

20

60

P

0.5

0.5

Therefore, the customer's expectation of the reward amount is (yuan).

(2) According to the budget of the mall, each customer will be rewarded with 60 yuan on average. So let's find a possible plan first, and predict 60 yuan. For the case that the par value consists of 10 yuan and 50 yuan, if we choose the scheme of (10,10,50), we expect it because 60 yuan is the maximum value of the sum of par values. If the scheme of (50, 50, 50, 10) is selected, because 60 yuan is the minimum value of the sum of face values, the mathematical expectation cannot be 60 yuan, so the possible scheme is (10, 10, 50, 50), which is recorded as scheme 1. For the face value of 20 yuan and 40 yuan, 20), so the possible scheme is (20, 20, 40, 40), which is recorded as Scheme 2. The following is an analysis of two schemes: for scheme 1, that is, scheme (10, 10, 50, 50), assuming that the reward won by the customer is, then

20

60

100

Expectation and variance of.

For Scheme 2, that is, Scheme (20, 20, 40, 40), let the customer's reward be, then the allocation table is.

40

60

80

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