Assignment on What is Probability Explain?.

Probability 1
PROBABILITY
by Student’s Name
Code + Course Name
Professor’s Name
University Name
City, State
Date

Probability 2
Probability
1. Coin Flips
A – heads atleast 4
B – heads equal to tails = 3
C – consecutive 4 heads
Pr ( H ) =1
2 ∧Pr ( T ) =1
2
Pr ( A )=Pr ( HHHH )∨Pr ( HHHHH )∨Pr ( HHHHHH ) orPr ( HHHHT ) orPr ( HHHHTT ) orPr ( HHHHTH )
¿ 1
24 + 1
25 + 1
26 + 1
25 + 1
26 + 1
26
¿ 11
64
Pr ( B ) =Pr ( HHH )=Pr ( TTT )
¿ 1
2
3
= 1
8
Pr ( C )=Pr ( HHHH )
¿ ( 1
2 ) 4
= 1
16
Pr ( A /B ) = Pr ( A ) Pr ( A ∩B )
Pr ( B )
¿
11
64 × 1
8
1
8
¿ 11
64

Probability 3
Pr (C / A)= Pr ( C ) Pr ( C ∩ A )
Pr ( A )
¿
1
16 × 1
16
11
64
= 1
44
2. Probability of 5 before 7
A – d1 + d2 = 7
B – d1 + d2 = 5
C - A ∪B
Pr ( 1 , … , 6 )= 1
6
The first appearance of A = 1 and 6 or 2 and 5 or 3 and 4 or 4 and 3
While first appearance of B = 1 and 4 or 2 and 3 or 3 and 2 or 4 and 1
A ∪B=(1,6)(1,5)(1,4 )(2,3)(2,4)(2,5)(3,2)(3,3)(3,4 )(4,1)(4,2)(4,3)
Pr ( A )= 1
36 + 1
36 + 1
36 + 1
36 = 1
9
Pr ( B ) = 1
36 + 1
36 + 1
36 + 1
36 = 1
9
Pr ( C )= 12
36
Pr ( C ∩ A )= 4
36
Pr (A /C)= Pr ( A)Pr ( C ∩ A )
Pr (C)

Probability 4
¿
1
9 × 1
9
12
36
= 1
27
3. Four-Door Monte Hall
With four doors, probability of;
Goat door opened Pr ( G ) = 3
4
Car door opened Pr ( C )= 1
4
i is the door closed but marked
j is a door with goat
i. Probability of winning is Pr ( C )=Pr ( i ) = 1
4
ii. Probability of winning given {j} is;
Pr ( C /{i }) implies doors are now 3
Pr (C)= 1
3
iii. Probability of winning given 2 doors,
Pr (C / { i, j } )= 1
2
4. Estimating Genetic Diseases
Pr ( C )= 1
25 , Pr (CC )= 1
4 , Pr ( C' )= 24
25 , Pr ( C C' ) = 3
4

Probability 5
Where, C – Carrier parent
CC – Carrier child
i. Probability two uniformly-chosen healthy people have child with CF
Pr ( CF )=Pr ( C ,CC )∧Pr ( C ,CC )
¿ 1
25 × 1
4 × 1
25 × 1
4 = 1
1000
ii. Probability two randomly – chosen parents have a non – carrier child
¿ Pr ( C ,C C' )∧Pr ( C , C C' )
¿ 1
25 × 3
4 × 1
25 × 3
4 = 9
1000
iii. Probability a carrier parent and a randomly chosen parent have a CF child
for a carrier parent , Pr (CC )= 1
2
for a randomly chosen parent , Pr ( C ) = 1
25 ∧Pr ( CC ) = 1
2
Pr ( CF )=Pr ( CC ) andPr ( C , CC )
¿ 1
2 × 1
25 × 1
2 = 1
100
iv. The probability that a carrier parent and randomly – chosen parent have a carrier
child
for a carrier parent , Pr (CC )= 1
2 ,∧Pr ( CC ' )= 1
2
for a randomly chosen parent , Pr ( C ) = 1
25 ∧Pr ( CC )= 1
2 ,∧Pr ( CC ' ) =1
2

Probability 6
¿ Pr ( CC )∧Pr ( C C' )∨Pr ( C C' ) ∧Pr ( CC )
¿ 1
100 + 1
100 = 1
50
v. The probability baby has CF from two uniformly chosen healthy people
Pr ( CF
CC )= Pr ( CF ) Pr (CF ∩ CC )
Pr (CC )
¿ Pr ( CF )
¿ 1
25 × 1
4 × 1
25 × 1
4 = 1
1000
5. Sampling With and Without Replacement
Probability of a cider ¿ 2
n
With replacement
From the geometric mean, E ( X ) = 1
p
¿ 1
( 2
n )
¿ n
2
Without replacement
We have 2
n
Then n−2
n × 2
n−1 , n−2
n × n−3
n−1 × 2
n−2 …
E ( X ) =2
n (1+2 ( n−2
n )+3 ( n−2
n )2
+… )=¿

Probability 7
¿ 2
n (2 ( n+1 )
n )
¿ n+1
3
6. Finding a Healthy Subring
1. E( X)
We have people = n, s = sick people, ps =probability of sick person
ph= probability of healthy person such that ps + ph=1
ps = s
n
Since n=k <100 intergers and k=s,
X Bin (k , ps )
ps = s
k
thus E ( X ) =k ps
2.
If n=70, s=39
We have, Pr ( ps )= 39
70 and Pr ( p¿¿ h)= 31
70 ¿

Probability 8
From a sample of n= 7, we expect Pr ( ps )= 39
70 × 7=4 ¿ be sick
¿ Pr ( ph ) = 31
70 ×7=3 ¿ be healthy
However, since the sample is random, a consecutive sample p0 , p1 , … , p6can have a
probability Pr ( ps ) ≤ 3
7 ∧Pr ( ph )≥ 4
7 while most of the sick people will be on the