Assignment about What is Probability?

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 𝑎𝑛𝑑 Pr(𝑇) = 1
2
Pr(𝐴)
= 𝑃𝑟(𝐻𝐻𝐻𝐻) 𝑜𝑟 𝑃𝑟(𝐻𝐻𝐻𝐻𝐻)𝑜𝑟 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)
=
1
9 × 1
9
12
36
= 1
27

Probability 4
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(𝐶𝐶′ ) = 3
4
Where, C – Carrier parent
CC – Carrier child
i. Probability two uniformly-chosen healthy people have child with CF

Probability 5
Pr(𝐶𝐹) = Pr(𝐶, 𝐶𝐶) 𝑎𝑛𝑑 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, 𝐶𝐶′ )𝑎𝑛𝑑 Pr(𝐶, 𝐶𝐶′ )
= 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 𝑎𝑛𝑑 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 , 𝑎𝑛𝑑 Pr(𝐶𝐶′) = 1
2
𝑓𝑜𝑟 𝑎 𝑟𝑎𝑛𝑑𝑜𝑚𝑙𝑦 𝑐ℎ𝑜𝑠𝑒𝑛 𝑝𝑎𝑟𝑒𝑛𝑡,
Pr(𝐶) = 1
25 𝑎𝑛𝑑 Pr(𝐶𝐶) = 1
2, 𝑎𝑛𝑑 Pr(𝐶𝐶′) = 1
2
= Pr(𝐶𝐶) 𝑎𝑛𝑑 Pr(𝐶𝐶′ ) 𝑜𝑟 Pr(𝐶𝐶′ ) 𝑎𝑛𝑑 Pr(𝐶𝐶)
= 1
100+ 1
100= 1
50

Probability 6
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
+ ⋯ ) =
= 2
𝑛( 2(𝑛 + 1)
𝑛 )
= 𝑛 + 1
3

Probability 7
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 𝑖𝑛𝑡𝑒𝑟𝑔𝑒𝑟𝑠and k=s,
𝑋~𝐵𝑖𝑛(𝑘, 𝑝𝑠)
𝑝𝑠 = 𝑠
𝑘
𝑡ℎ𝑢𝑠 𝐸(𝑋) = 𝑘𝑝𝑠
2.
If n=70, s=39
We have, Pr(𝑝𝑠) = 39
70 and Pr (𝑝ℎ) = 31
70
From a sample of n= 7, we expect Pr(𝑝𝑠) = 39
70 × 7 = 4 𝑡𝑜 𝑏𝑒 𝑠𝑖𝑐𝑘
𝑎𝑛𝑑 Pr(𝑝ℎ) = 31
70× 7 = 3 𝑡𝑜 𝑏𝑒 ℎ𝑒𝑎𝑙𝑡ℎ𝑦
However, since the sample is random, a consecutive sample 𝑝0, 𝑝1, … , 𝑝6can have a
probability Pr(𝑝𝑠) ≤ 3
7 𝑎𝑛𝑑 Pr(𝑝ℎ) ≥ 4
7 while most of the sick people will be on the