Hypergeometry

## 2845535

https://math.stackexchange.com/questions/2845535

Probability of sixth ball to be white

A box contains 7 identical white balls and 5 identical black balls. They are to be drawn randomly one at a time without replacement until the box is empty. Find the probability that the 6th ball drawn is white, while before that exactly 3 black balls are drawn.

*Principle and techniques in combinatorics by Chen Chuan Chong, Ch 1, quest. 24*

g
o
o
d

a
l
l

=

2

W
⋅

3

B

5

(
W
B
)

W

W
B

4

W
⋅

2

B

6

(
W
B
)

{\displaystyle {\frac {good}{all}}={\frac {{\mathit {2}}W\cdot {\mathit {3}}B}{{\mathit {5}}(W+B)}}{\frac {W}{W+B}}{\frac {{\mathit {4}}W\cdot {\mathit {2}}B}{{\mathit {6}}(W+B)}}}

Above is the species presentation. The e.g.f's are

w

2

2
!

{\displaystyle w^{2} \over 2!}

w

{\displaystyle w}

w

4

4
!

{\displaystyle w^{4} \over 4!}

b

3

3
!

{\displaystyle b^{3} \over 3!}

1

{\displaystyle 1}

b

2

2
!

{\displaystyle b^{2} \over 2!}

(
b
w

)

5

5
!

{\displaystyle (b+w)^{5} \over 5!}

b
w

{\displaystyle b+w}

(
b
w

)

6

6
!

{\displaystyle (b+w)^{6} \over 6!}

g
o
o
# d

w

2

b

3

2
!
.3
!

w

w

4

b

2

4
!
.2
!

{\displaystyle good={\frac {w^{2}b^{3}}{2!.3!}}{w}{\frac {w^{4}b^{2}}{4!.2!}}}

 ,

a
l
# l

(
w
b

)

5

5
!

(
w
b
)

(
w
b

)

6

6
!

{\displaystyle all={(w+b)^{5} \over 5!}(w+b){(w+b)^{6} \over 6!}}

Now we compare the coefficients of

w

7

b

5

7
!
5
!

{\displaystyle w^{7}b^{5} \over 7!5!}

 in the two e.g.f.'s above and we get:

1050
5544

=

25
132

{\displaystyle {\frac {1050}{5544}}={\frac {25}{132}}}

## 3484751

https://math.stackexchange.com/questions/3484751/ Combinations when repetitions are allowed

Let there be 10 gifts of 3 kinds (A,B,C); of which we have 5A,3B,2C. If we are supposed to select 7 gifts out of these, what is the probability that we pick a combination of 4A,2B,1C ?

g
o
o
d

a
l
l

=

4

(
A
)
⋅

2

(
B
)
⋅
C

7

(
A
B
C
)

A
⋅
B
⋅
C

3

(
A
B
C
)

{\displaystyle {\frac {good}{all}}={\frac {{\mathit {4}}(A)\cdot {\mathit {2}}(B)\cdot C}{{\mathit {7}}(A+B+C)}}{\frac {A\cdot B\cdot C}{{\mathit {3}}(A+B+C)}}}

a

5

5
!

{\displaystyle a^{5} \over 5!}

a

4

4
!

{\displaystyle a^{4} \over 4!}

a

1

1
!

{\displaystyle a^{1} \over 1!}

b

3

3
!

{\displaystyle b^{3} \over 3!}

b

2

2
!

{\displaystyle b^{2} \over 2!}

b

1

1
!

{\displaystyle b^{1} \over 1!}

c

2

2
!

{\displaystyle c^{2} \over 2!}

c

1

1
!

{\displaystyle c^{1} \over 1!}

c

1

1
!

{\displaystyle c^{1} \over 1!}

a

5

5
!

b

3

3
!

c

2

2
!

{\displaystyle {a^{5} \over 5!}{b^{3} \over 3!}{c^{2} \over 2!}}

(
a
b
c

)

7

7
!

{\displaystyle (a+b+c)^{7} \over 7!}

(
a
b
c

)

3

3
!

{\displaystyle (a+b+c)^{3} \over 3!}

remark that

c
o
e
f
f

o
f

a

5

5
!

b

3

3
!

c

2

2
!

=
c
o
e
f
f

o
f

a

6

6
!

b

2

2
!

c

2

2
!

=
# ⋯
c
o
e
f
f

o
f

a

10

10
!

b

0

0
!

c

0

0
!

=

(

10
7

)

{\displaystyle coeff\ of\ {a^{5} \over 5!}{b^{3} \over 3!}{c^{2} \over 2!}=coeff\ of\ {a^{6} \over 6!}{b^{2} \over 2!}{c^{2} \over 2!}=\cdots =coeff\ of\ {a^{10} \over 10!}{b^{0} \over 0!}{c^{0} \over 0!}={10 \choose 7}}

## 2845039

https://math.stackexchange.com/questions/2845039

A standard deck of cards is shuffled and dealt. Find the probability that the last king appears on the 48th card?

Let X be the sort of kings and Y the sort of non-kings. We are forming a partition 47+1+4 using labelled cards.

then the species for the good/all configurations (with a king in the special position 48) is

g
o
o
d

a
l
l

=

E

47

(
X
Y
)

E

47

(
X
Y
)

X

X
Y

E

4

(
Y
)

E

4

(
X
Y
)

{\displaystyle {\frac {good}{all}}={\frac {E\_{47}(X+Y)}{E\_{47}(X+Y)}}{\frac {X}{X+Y}}{\frac {E\_{4}(Y)}{E\_{4}(X+Y)}}}

The e.g.f's are

g
o
o
# d

(
x
y

)

47

47
!

x

y

4

4
!

{\displaystyle good={\frac {(x+y)^{47}}{47!}}x{\frac {y^{4}}{4!}}}

a
l
# l

(
x
y

)

47

47
!

(
x
y
)

(
x
y

)

4

4
!

{\displaystyle all={(x+y)^{47} \over 47!}(x+y){(x+y)^{4} \over 4!}}

We now compare the coefficients of

X

4

4
!

Y

48

48
!

{\displaystyle {\frac {X^{4}}{4!}}{\frac {Y^{48}}{48!}}}

 in the above expressions - since there are 4 kings and 48 non-kings.

g
o
o
d

a
l
l

=

778320
12994800

≈
0.06

{\displaystyle {\frac {good}{all}}={\frac {778320}{12994800}}\approx 0.06}

Working with GF :

We have two kinds of cards, kins and no-kings, to be placed in 52 specified slots.

(
X
Y

)

47

⋅
X
⋅

Y

4

{\displaystyle (X+Y)^{47}\cdot X\cdot Y^{4}}

The GF for all configurations is:

(
X
Y

)

47

⋅
(
X
Y
)
⋅
(
X
Y

)

4

{\displaystyle (X+Y)^{47}\cdot (X+Y)\cdot (X+Y)^{4}}

g
o
o
d

a
l
l

=

16215
270725

≈
0.06

{\displaystyle {\frac {good}{all}}={\frac {16215}{270725}}\approx 0.06}