maths probability problems

1.
No. of combinations without restriction
= 52C5


(a)
For the event to happen :
Select 1 suit from the 4 suits. (4C1)
Select 5 cards from the 13 cards of the suit. (13C5).

The required probability
= 4C1 x 13C5 / 52C5
= (4!/3!1!) x (13!/8!5!) / (52!/47!5!)
= 33/16660


(b)
For the event to happen :
Select 2 kings for the 4 kings. (4C2)
Select 3 values from the 13 values. (13C3)
Select 'heart' or 'diamond' for each of the 3 values. (2C1³)

The required probability
= 4C2 x 13C3 x 2C1³ / 52C5
= (4!/2!2!) x (13!/10!3!) x 8 / (52!/47!5!)
= 22/4165


(c)
For the event to happen :
(1) Select 1 value from the 13 values. (13C1)
(2) Select 2 black cards and 1 red card from the 4 cards of the value in (1). (2C2x 2C1)
(3) Select 1 value from the rest 12 values (12C1)
(4) Select the 2 black cards from the 4 cards of the value in (3). (2C2)

The required probability
= 13C1 x 2C2 x 2C1x 12C1 x 2C2 / 52C5
= 13 x 1 x 2 x 12 x 1 / (52!/47!5!)
= 1/8330

(b) 有機會抽 3 隻 KINGS……