Module 4. Assignment:

Question A:


 

 

 

 

A1: What is the probability of event A?

P_Total = 10 + 20 + 20 + 40 = 90

P(A) = (10 + 20)/90 = 30/90 = 3/9 = 1/3 = 0.333333... 

A2: What is the probability of event B?

P(B) = (10 + 20)/90 = 30/90 = 3/9 = 1/3 = 0.333333...

A3: What is the probability of event A or B?

P(A ∩  B) = 10/90 = 1/9 = 0.1111

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/3 + 1/3 - 1/9 = 5/9 = 0.5556

A4: What is the probability of P(A or B) = P(A) + P(B)?

P(A or B) = P(A ∪ B)

P(A ∪ B) = P(A) + P(B) only if they're mutually exclusive.

However, since they are not mutually exclusive, that means they overlap. Therefore, the probability of it happening is zero.

Question B:

What is the probability that it will rain on the day of Jane's wedding?

P(rain) = 5/365

P(forecast rain | rain) = 0.9

P(forecast rain | no rain) = 0.1

P(rain | forecast rain) = ((5/365)(0.9))/((5/365)(0/9)) + ((360/365)(0.1)) = 0.111

B1: 

Based on the evidence, we can announce with complete certainty that this answer is true.

B2:

The reason why this answer is true is because the rainfall is very rare (only 5 days/year), and most rain forecasts happen on days where rainfall doesn't occur. This is still the case even if the forecast is accurate as there are so many non-rain days. Therefore, the low rainfall rate keeps the probability low.

Question C:

C1:  What is the probability of operating on 10 patients successfully with the traditional method?

The complication frequency is at 20%, therefore:

n = 10

p = 0.20

We need zero complications, so the code shall be:

dbinom(0, size = 10, prob = 0.2)

When ran in the R Console, we get:

0.1073742 = 10.74% 


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