[20分] 星期二測驗，緊急求救:probability問題

1a: By law of total probability: Pr(A & B) = Pr(A) + Pr(B) – Pr(A or B) = 1/9
1b: Pr(C|B) = Pr(B & C) / Pr(B) = 4/9
1c:
By law of total probability: Pr(A & C) = Pr(A) + Pr(C) – Pr(A or C) = 1/6
Pr(A|C) = Pr(A & C) / Pr(C) = 5/12

2: X ~ Binomial (n=8, p=0.3)
2a: Pr(X=3) = 8C3 x 0.33 x (1 – 0.3)8-3 = 56 x 0.027 x 0.16807 = 0.2541
2b: Pr(2<=X<=4) = Pr(X=2) + Pr(X=3) + Pr(X=4) = 8C2 x 0.32 x (1 – 0.3)8-2 + 0.2541 + 8C4 x 0.34 x (1 – 0.3)8-4 = 28 x 0.09 x 0.117649 + 0.2541 + 70 x 0.0081 x 0.2401 = 0.6867

3a: Pr(at least 2 seeds would germinate) = 1 – Pr(no seed would germinate) – Pr(1 seed would germinate) = 1 – (1 – 7/9)9 – 9 x (7/9) x (1 – 7/9)8 = 0.999957049 ≈ 1.0

4b:
Pr(a class with exactly 8 students having mobile phone) = 15C8 x 0.78 x (1 – 0.7)15 – 8 = 6435 x 0.0576 x 0.0002187 = 0.0811
Pr(class number 4 is the first class to have exactly 8 students having mobile phone) = (1 – 0.0811)3 x 0.0811 = 0.0629

5: X ~ Poisson (l=4)
5aII: Pr(X=3) = 43 x e-4 / 3! = 0.1954
5b: variance of X = mean of X = l = 4

2010-03-29 14:15:42 補充：
your suggested answer for no. 3 may be wrong

1a
P(A and B) = P(A) + P(B) - P(A or B) = 1/3 + 1/2 - 13/18 = 2/18 = 1/9

1b
P(C and notB) = P(C) - P(C and B) = 2/5 - 2/9 = 8/45

1c
P(A and notC) = P(A) - P(A and C) = P(A) - [P(A)+P(C)- P(A or C)]= 1/3 - (1/3 + 2/5- 17/30) = 1/3 -1/6 = 1/6

2. Forgot how to do....sorry

3a
P( at least 2 germinate) = 1 - P(no germinate) - P( 1 germinate)
= 1 - (2/9)^9 - (9C1)x(2/9)^8 x (7/9)



too late already...幫住咁多先~

1. It is given that A, B and C are three events with P(A) = 1/3, P(B) = 1/2, P(C) = 2/5, P(A OR B) = 13/18, P(A or C) = 17/30 and P(B and C) 2/9
(a)P(A and B)
=P(A)xP(B)
=1/3x1/2
=1/6
(b)P(C∣B)
=(2/9)/(1/2)
=4/9
(c)P(A∣C)


3. The chance that a seed will germinate is 7/9. If 9 seeds are sown, what is the personality that
(a) At least 2 seeds will germinate?
1-(2/9)^9-(7/9)(2/9)^8 =1.00