# SOLUTION: American Academy of English Binomial Distribution and Probability Problems

Hello, Please confront solid the corrected answers of these problems.I possess corrected the tasks 1, 2, 3, 4, 7, 8, 10, 11Please possess a face. Thanks :))

My ID: 18f18336
5 other IDs: 18s19456, 19f19231, 18f18247, 18f18332, 18f18458
Hence, Aggregate aggregate = (7)*(6) = 42 = [13 eras ‘1’, 3 eras ‘2’, 5 eras ‘3’, 3 eras ‘4’, 2
times ‘5’, 2 eras ‘6’, 1 eras ‘7’, 10 eras ‘8’, 3 eras ‘9’]
And Aggregate scholarship = (1)*(6) = 6 [5 eras ‘f’, 1 era ‘s’]
Since, each card contains either a sum or a epistle for-this-reason there would be a aggregate of 42 + 6
= 48 cards in the box
Let incident A: is selecting a card that contains a sum
And incident B: is selecting a card that contains a sum
a)
42

i.
ii.

Probability of a sum P(A) = 42+6 = 0.875
Probability of a epistle P(B) = (1 – 0.875) = 0.125

i.

P(twain scholarship) = 48 𝑥

b)

ii.

6

6

6

48
5

P(twain scholarship) = 48 𝑥

48

= 0.0156
= 0.0130

Don’t patronage inequitable curfew
17

Supports inequitable curfew
28

a) Calculate the likelihood of beings who patronage a inequitable curfew
28
P = 28+17 = 0.622

b) Calculate the likelihood of beings who do not patronage a inequitable curfew
17
P = 28+17 = 0.377

Male student

Female student

Gender
Face-to-face learning
Online learning
No difference
Total

Male
9
5
1
15

Female
9
5
1
15

a) The student does fancy online le...