Probability Assignment: Statistics and Probability Concepts Explored

This assignment delves into various probability concepts through a series of problems. It begins with coin flip scenarios, analyzing the probability of getting at least four heads, the probability of an equal number of heads and tails, and the probability of consecutive heads. The assignment then explore...Read More

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