maths probability

1)
Since the sum is even, therefore the two number must be either odd + odd or even + even
the total combinations for even sum =3*3+3*3=18
since there cannot be a 4 in odd+odd case, so we only consider the even+even case
the combinations at least a 4=(total combinations of even+even) - (both two nos are not 4)
=3*3-2*2=5
there P(at least a 4/sum of the two nos is even)=5/18

2)
the total combinations = 6*5*4*3*2*1
Let N(x) be the no of ways to put balls into x box.
Let us consider we first put balls into box 1 and box 6
the combinations that A is not in box 1 and B is not in Box 6
=N(1 given that it is not ball A)*N(6 given that it is not ball B)*N(2)*N(3)*N(4)*N(5)
=[N(1 given that it not ball A and not ball B)*N(6 given that it is not ball B) +N(1 given that it is ball B)*N(6 given that it is not ball B)]*N(2)*N(3)*N(4)*N(5)
=[4*4+1*5]*4*3*2*1=(21)*4*3*2*1
therefore the the probability that ball A is not in box 1 and ball B is not in box 6 =(21)*4*3*2*1/6*5*4*3*2*1
=21/30=7/10


2007-11-06 09:33:15 補充：
為甚麼是6!而不是6^6，這是因為將波放入第一個箱有６個組合，將波放入第二個箱時，就只餘下５個球１，所以就只有５個組合，如此類推，總組合是６！

Q1Two fair dice are tossed. It is known that the sum of the two numbers obtained is an even number. Fing the probability that at least a '4' is obtained.
　　１　　２　　３　　４　　５　　６
１　／　　　　　／　　　　　／
２　　　　／　　　　　Ｘ　　　　　／
３　／　　　　　／　　　　　／
４　　　　Ｘ　　　　　Ｘ　　　　　Ｘ
５　／　　　　　／　　　　　／
６　　　　／　　　　　Ｘ　　　　　／
/ : 代表 sum of the two numbers obtained is an even number (包括 X 位)
X : 代表 at least a '4' is obtained
/ : 有 18 個
X : 有 5 個
p(at least a '4' ! even sum) = 5/18
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Q2Six balls A, B, C, D, E and F are put into six boxes numbered 1, 2, 3, 4, 5, and 6. If only one ball can be put into each box, what is the probability that ball A is not in box 1 and ball B is not in box 6?
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