Mod 4: Probability Theory

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Set environment

env <- c("dplyr")
lapply(env, library, character.only = 1)

Contingency Table

Create table

data1 <- matrix( c(10,20,20,40), nrow=2, byrow=1,
                 dimnames=list( c("A","A1"), c("B","B1") ) )
table1 <- as.table( data1 )
table1

##     B B1
## A  10 20
## A1 20 40

Calculate relative frequencies

propTable <- prop.table(table1)
propTable

##            B        B1
## A  0.1111111 0.2222222
## A1 0.2222222 0.4444444

Find probabilities for ‘A’ and ‘B’

• P(A) : sum the table row
  pA <- propTable %>% rowSums() %>% .["A"] = 0.3333333
  

• P(B) : sum the table column
  pB <- propTable %>% colSums() %>% .["B"] = 0.3333333
  

• P(A and B) : is already on the rel freq table as \(P(A) \cap P(B)\) but we can still calculate
  pA * pB = 0.1111111
  

• P(A or B) : P(A) + P(B) - P(A and B)
  pA + pB - (pA * pB) = 0.5555556
  

• P(A or B) disjoint : P(A) + P(B)
  pA + pB = 0.6666667
  
  
  

Bayes Theorem

Jane is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn’t rain, he incorrectly forecasts rain 10% of the time.

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

Discussion

Calculate the probabilities we know. ( R - rain, NR - no rain, F - weatherman forecast )

• Relative
  • P(R) : pR <- 5/365 = 0.0136986
    
  • P(NR) : pNR <- 360/365 = 0.9863014
    
• Conditional
  • P(F|R) : 0.9 pFR <- 0.9
    
  • p(F|NR) : 0.1 pFNR <- 0.1
    
    

Proportional probability would suggest that there’s a 0.01 or 1.37% chance that it will rain on Jane’s wedding day.

Given the weatherman’s prediction history, we can use Bayes to update our initial probability - the prior. We now want to predict the probability of rain based on the new forecast information - the posterior : \(P(R|F)\)

Using Bayes : \(\large P(R|F) = \frac{P(F|R) \cdot P(R)}{P(F)}\) = \(\large \frac{P(F|R) \cdot P(R)} {P(F|R) \cdot P(R) + P(F|NR) \cdot P(NR)}\)

Substituting results in : \(\large \frac{0.9 \cdot 0.0136986} {0.9 \cdot 0.0136986 + 0.1 \cdot 0.9863014}\) = 0.1111111

Interpretation:

• The weatherman’s forecast increased our probability of rain from 1.37% to 11.11%. A big increase but overall, there’s a low probability of rain on Jane’s wedding day.
  
• While the weatherman’s rain prediction rate is high, the fact that it only rains a few days per year, significantly lowers his daily probability accuracy. Even if the weatherman had an accurate rain forecast rate of 99%, the probability of rain on the wedding day would still be only about 0.58 or 58% - slightly better than a coin toss.
  
  

GitHub

Related file(s) can be found at Git Me