1+3+5=9, in which p (3,3) = 6 kinds; 1+4+4=9, and there are three kinds of p (3,3)/p (2,2) = 3;
2+2+5=9, there are three kinds (enumeration method); 2+3+4=9, in which P (3,3) = 6 kinds;
3+3+3=9, only 1 species; Total probability of 3 out of 5 (repeatable): 5 * 5 * 5 =125;
∴ The probability that the sum of three digits equals 9 is: (6+3+3+6+1)125 =19/125.
2. The different situations of taking two days off seven days a week are: c (7 7,2) = 21,
Then, the situation that three people work and rest at the same time is only 2 1.
The probability that everyone chooses a certain situation is 1/2 1, which is obviously independent of each other.
∴ The probability of working and resting at the same time is: 21(21* 21* 21) =1/441.
3. (Binding Method and Insertion Method)
There are P(3, 3) different situations in which three students in a class are arranged together.
As a whole (binding), these three students are all arranged with five students from other classes, and there are P (6,6) kinds.
Then two students from Class Two are inserted into seven gaps formed by the above six students (so they are not together), and there are P (7,2) kinds.
The total probability is P( 10, 10).
∴ Find the probability: p (3,3) * p (6,6) * p (7,2)/2)/p (10/0,10) =120.
4. There are two groups A and B, and the teams of 10 are numbered from 1 to 10 from strong to weak. The two strongest teams are: 1 and 2, and the rest is the draw.
The probability that each team draws a certain group is 1/2, and which group each team draws is independent of each other.
A.)P (the top two teams are divided into different groups)
=P( 1 smoke a, 2 smoke B)+ P( 1 smoke b, 2 smoke a) =1/2 *1/2+1*/kloc-0 =
B.)P (the two strongest teams are in the same group)
=P( 1, 2 drawing A)+ P( 1, 2 drawing b) =1/2 *1/2 =1/.
PS。 The topic itself is not complicated, but quite interesting. I wish you progress in understanding probability and mastering common methods!