F.5 math's on probability

1.

We can know the probability only while we don't know the number, so we let
unknown will be easier.

Let the number of children be n.

P( at least a son and a daughter) > 0.9
1 - P(all boys) - P(all girls) > 0.9
1- (1/2)^n -(1/2)^n > 0.9
2(1/2)^n < 0.1
(1/2)^n <0.05
n log(1/2) <log 0.05
-0.301 n <-1.301
n > 4.32
So the number of children is 5.


2.
(a)
P(either or both of them is absent) = 1 - P(all present)
= 1- (1-0.05)(1-0.06)
= 0.107
(b)
P(at least one of them comes to the class)= 1 - P(all absent)
= 1- (0.06)(0.05)
= 0.997
(c)
P(only one of them comes to the class)= P(Tom comes and Peter not comes)
+ P(Peter comes and Tom not comes)
= 0.06(1-0.05) + 0.05( 1- 0.06)
= 1.04

Alternatively,
(c)
By (a),(b)
P(either or both of them is absent) = P(both absent)
+P(only one of them comes to the class)
0.107 = 0.003
+P(only one of them comes to the class)
P(only one of them comes to the class) =0.104

2013-03-02 18:18:31 補充：
D位亂左小小,果D下面有等號係方程

2013-03-02 18:19:40 補充：
1+
Alternatively,
(c)
係方程