Module 4. Assignment

A. Based on Table 1 What is the probability of:

	B	B1
A	10	20
A1	20	40

A1. Event A =   P(A) = 30/90 = 1/3

A2. Event B? =  P(B) = 30/90 = 1/3

A3. Event A or B = P(A or B) =  P(A) + P(B) - P(A and B)

                                            1/3 + 1/3 - 1/9 = 5/9

A4. P(A or B) = P(A) + P(B) =   False, because P(A or B) = 5/9 and P(A) + P(B) = 2/3

B1. True

B2. We  are given the following: 

In terms of probabilities, we know the following:
P( A1 ) = 5/365 =0.0136985 [It rains 5 days out of the year.]
P( A2 ) = 360/365 = 0.9863014 [It does not rain 360 days out of the year.]
P( B | A1 ) = 0.9 [When it rains, the weatherman predicts rain 90% of the time.]
P( B | A2 ) = 0.1 [When it does not rain, the weatherman predicts rain 10% of the time.]
We want to know P( A1 | B ), the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below.

We want to know P( A1 | B ), the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman.

The formula is:

P( A1 | B) = P( B | A ) x P( A1 ) / P( A1) x P(B | A1)  x  P(A2) x P( B | A2 ) 

The numerator is P ( B | A) x P( A1) because it is a joint probability of both rain and the weatherman predicting rain, is represent by:

P( B | A ) x P( A1 ) 

The denominator is the total probability that the weatherman predicted rain, and it does rain, and it doesn't rain, which is represented by: 

P( B | A1 ) x  P( A1) x P( B | A2 ) x P( A2 ) 

Using the given, we plug everything in: 

P( A1 | B ) = 0.9 x 0.0136985 / (0.9 + 0.0136985) + (0.1 x 0.9863014)

= 0.01232865 / 0.11095879

=0.111

= 11.1%

C.

> dbinom(10, size=10, prob=0.8)
[1] 0.1073742