Question

A bridge hand is made up of 13 cards from a deck of 52. find the probability that a hand chosen at random contains at least 3 kings kings.

Answer 1

1.
In total there are C(52, 13) ways that we can pick a hand, that is $\frac{52!}{13!39!}$


2.
P(a hand contains at least 3 kings)
                =P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

3.
first let's find P(a hand contains exactly 3 kings):

P(a hand contains exactly 3 kings)
              =n(a hand contains exactly 3 kings)/C(52, 13)
              
n(a hand contains exactly 3 kings)=C(4, 3)*C(48, 10)

where C(4,3) is the total number of ways we can pick 3 out of 4 kings,

C(48, 10) is the number of picking 10 letters to complete a hand, out of the 52-4=48 non-king cards.

so P(a hand contains exactly 3 kings)=[C(4, 3)*C(48, 10)]/C(52, 13)

4. with the same reasoning as in step 3:

P(a hand contains 4 kings)=n(a hand contains 4 kings)/C(52, 13)

                          = [C(4, 4)*C(48, 9]/C(52, 13)


5. 

P(a hand contains at least 3 kings)
                =P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

=[C(4, 3)*C(48, 10)]/C(52, 13)+ [C(4, 4)*C(48, 9)]/C(52, 13)

=$\frac{C(4, 3)*C(48, 10)+C(4, 4)*C(48, 9)}{C(52, 13)}$

=$\frac{4* \frac{48!}{10!38!} + \frac{48!}{9!39!}}{ \frac{52!}{13!39!} }$

simplify by 38! in the denominators and 48! in the numerators :

$\frac{4* \frac{1}{10!} + \frac{1}{9!39}}{ \frac{52*51*50*49}{13!39} }$

now simplify by 9! in all denominators:

$\frac{ \frac{4}{10}+ \frac{1}{39} }{ \frac{52*51*50*49}{13*12*11*10*39}}$


$\frac{ 0.426 }{ 9.7} =0.044$