It should be n = p n=P( 10/0,5) = 30240,
It means to select 5 tickets from 10 to arrange.
Your m(A 1)=2520 and m(A2)=5040 are all right.
M(A3) is wrong. You should take P (3 3,2) instead of C (3 3,2) and get two. You haven't allocated them yet!
m(A3)=C(4，2)×P(3，2)×7×6 = 6×6×7×6 = 15 12
So the probability is: (2520+5040+1512)/30240 = 3/10.
Second, explain your teacher's methods as follows:
Assuming that all 10 people are taken, the total number of taking methods is P( 10, 10)= 10!
In the process of taking it, it doesn't matter much because of everyone's order. We can number the fifth person as 1 and let him choose first.
Obviously, there are three possibilities for his choice.
So the number of basic events satisfying a is: 3× p (9,9) = 3× p (9!
Thirdly, since this problem is a conditional probability problem, we can also refer to the total probability formula in conditional probability.
Let Bi={ I got the admission ticket in the top 4}
Then p (B0) = (7× 6× 5× 4)/(10× 9× 8× 7) =1/6, and P(A|B0)=3/6= 1/2.
P(B 1)=C(4， 1)×P(3， 1)×7×6×5/( 10×9×8×7)= 1/2，P(A|B 1)=2/6= 1/3
P(B2)=C(4，2)×P(3，2)×7×6/( 10×9×8×7)= 10/3，P(A|B2)= 1/6
P(A|B3)=0
So p (a) = p (b0) p (a | b0)+p (b1) p (a | b1)+p (b2) p (a | b2)+p (b3) p (a | b3).
= 1/6× 1/2+ 1/2× 1/3+3/ 10× 1/6+0=3/ 10
Appendix: full probability formula/view/ACB1174469eae009581bec4.html.
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