. In recent years, droughts often occur in many areas of China, and artificial rainfall is often needed to combat drought. According to the weather forecast,

Solution: (1) The probability of no artificial rainfall for 5 days is P 1 = (

1

2

)3(

four

five

)2=

2

25

, ... (2 points)

Therefore, the probability that at least artificial rainfall is needed 1 day is 1-p 1=

92

100

=

46

50

=

23

25

..... (4 points)

(2) The value of x is 0, 1, 2, 3, 4, 5,

According to (1), the probability that artificial rainfall is not needed for 4 days is:

P(x=4)=(

1

2

)3C

1

2

four

five

×

1

five

+C

1

three

(

1

2

)3(

four

five

)2=

seven

25

, ... (5 points)

p(x = 3)= 1

C

2

three

(

1

2

)3(

four

five

)2+C

1

three

(

1

2

)3C

1

2

(

1

five

)(

four

five

)+(

1

2

)3(

1

five

)2=

73

200

, ... (6 points)

The probability of not needing artificial rainfall for 2 days is:

P(x=2)=C

2

three

(

1

2

)3(

1

five

)2+C

1

three

(

1

2

)3C

1

2

(

1

five

)×(

four

five

)+(

1

2

)3×(

four

five

)2=

43

200

, ... (7 points)

The probability that 1 day does not need artificial rainfall is: P(x = 1)= 1

1 1

200

, ... (8 points)

The probability that artificial rainfall is not needed for 0 days is: P(x=0)= (

1

2

)3(

1

five

)2=

1

200

, ... (9 points)

The distribution table of X days without artificial rainfall is: ... ……( 10/0 point)

x 0 1 2 3 4 5

P

1

200

1 1

200

43

200

73

200

seven

25

2

25

The expected number of days without artificial rainfall x is:

F 1=0×

1

200

+ 1×

1 1

200

+2×

43

200

+3×

73

200

+4×

seven

25

+5×

2

35

= 3. 1 ...( 12 points)