SOLUTION: American Academy of English Binomial Distribution and Probability Problems

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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 :))

Task 1
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

Task 2

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

Task 3
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...


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