概率問題....[容斥定理]

1) Same as nitro~
2) Same as myisland8132

Just think about it.

1 In class of 40 student, 22 of them have exempted Maths; 7 of them have exempted Eng whereas 15 of them have to study both subject. If student is picked at random from class, find the probability that
a exempted Maths
b exempted at least one subject
c exempted both subject
d exempted Eng but not Maths
(a)
22 of them have exempted Maths; 7 of them have exempted Eng whereas 15 of them have to study both subject
N(E or M)=40-15=25
ALSO
N(E or M)=N(E)+N(M)-N(E and M)
25=22+7-N(E and M)
N(E and M)=4
We get
N(E)=22-4=18
N(M)=7-4=3
the probability that exempted Maths
=22/40
=11/20
(b)
the probability that exempted at least one subject
=25/40
=5/8
(c)
the probability that exempted both subject
=4/40
=1/10
(d)
the probability that exempted Eng but not Maths
=3/40
2 Ann and Bob take turn to toss a coin until a "Head&q uot; turns up. The first one getting a "Head&q uot; wins. If Ann starts first, find the probability that Ann will win.
Consider Ann
when Ann getting a head, she wins
probability=1/2
when Ann gets a tail, Bob gets a tail, then Ann getting a head, she wins
probability=(1/2)^3
The probability that Ann will win is just equal to
1/2+(1/2)^3+(1/2)^5+....
=(1/2)/(1-1/4)
=1/2*4/3
=2/3



2006-12-10 03:55:13 補充：
N(E) represents the number of students exempted EnglishSimilar for others