Maths Probability

1.A=(total more than 9)
Let X be the total number of two rolled
Let Y be the number of first rolled
Let Z be the number of second rolled
P(A)=P(X=10)+P(X=11)+P(X=12)
=P(Y=4∩Z=6)+P(Y=5∩Z=5)+P(Y=6∩Z=4) +
P(Y=5∩Z=6)+P(Y=6∩Z=5)+
P(Y=6∩Z=6)
=3/36 + 2/36 + 1/36
= 1/6

2.B=(one even number showing)
P(B) = 1-P(both rolled are even)-P(both rolled are odd)
= 1-3/6*3/6 - 3/6*3/6
= 1/2

P(A∩B)
= P(Y=5∩Z=6) + P(Y=6∩Z=5)
=1/36+1/36
=1/18
P(A∩B^c) B^c is the negate of B
= P(Y=4∩Z=6)+P(Y=5∩Z=5)+P(Y=6∩Z=4)+P(Y=6∩Z=6)
=4*1/36
=1/9

i dont know Find A∩B' means A∩B or A∩B^c
so i had done both of them
hope can help you

2008-07-08 20:39:40 補充：
B= {one even number showing},
B^c=(no even number showing,all even number showing)

B'= one showing odd no.
∴A∩B' ={(5,5),(5,6),(6,5)}
∴The required prob=1/12