# 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 :))

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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