S5 Math Ex.

1.
(a)
B : a black sock
W : a white sock

There are 4B and 8W, totally 12 socks.

P(B in the first two draws)
= P(B) + P(WB)
= (4/12) + (8/12) x (4/11)
= (11/33) + (8/33)
= 19/33

(b)
P(B in the first three draws)
= [P(B) + P(WB)] + P(WWB)
= (19/33) + (8/12) x (7/11) x (4/10)
= (95/165) + (28/165)
= 41/55


2.
(a)
No. of selections in random
= 12C6

Select the 2 heavies students (2C2) and 4 other students(10C4).
No. of the required selections
= 2C2 x 10C4

The required probability
= 2C2 x 10C4 / 12C6
= 1 x (10!/4!6!) / (12!/6!6!)
= 5/22

(b)
Select the lightest student (1C1) and 5 other students (11C5).
No. of required selections
= 1C1 x 11C5

The required probability
= 1C1 x 11C5 / 12C6
= 1 x (11!/5!6!) / (12!/6!6!)
= 1/2

(c)
Select the heaviest student (1C1) and 5 other studentsexcluding the lightest student (10C5).
No. of the required selection
= 1C1 x 10C5

The required probability
= 1C1 x 10C5 / 12C6
= 1 x (10!/5!5!) / (12!/6!6!)
= 3/11


3.
(a)
M : a man
W : a woman
There are 6M and 5W, totally 11 persons.

No. of selections without restriction
= 11C4

To select 4M, no. of the required selections
= 6C4

The required probability
= 6C4 / 11C4
= (6!/4!2!) / (11!/4!7!)
= 1/22

(b)
To select 4M or 4W, no. of the required selections
= 6C4­ + 5C­4

The required probability
= (6C4 / 11C4) + (5C4/ 11C4)
= (1/22) + (5!/4!1!) / (11!/4!7!)
= (3/66) + (1/66)
= 2/33

(c)
To select 2M and 2W, no. of the required selections
= 6C2­ x 5C­2

The required probability
= 6C2­ x 5C­2 / 11C4
= (6!/2!4!) x (5!/2!3!) / (11!/4!7!)
= 5/11

(d)
To select 4M or (3M and 1W), no. of the required selections
= 6C4 + 6C3 x 5C1

The required probability
= (6C4 / 11C4) + (6C3x 5C1 / 11C4)
= (1/22) + (6!/3!3!) x 5 / (11!/4!7!)
= (3/66) + (20/66)
= 23/66