Data & Statistics

You may have known how to evaluate a definite integral or integrals bounded within limits. For example, consider the following:

We can estimate the value by integrating the function y = x² and applying the limits.

The function (within those bounds) is plotted below, and the required integral is the area under the curve.

Looking at the graph above, one can envision that the required area is the fraction of a unit square at which the relationship, y </= x² works. As we have done before, we will fill the unit square by selecting random numbers between 0 and 1 and collect the fraction that obeys the equation. This this:

itr <- 10000

xxx <- replicate(itr,0)
yyy <- replicate(itr,0)
counter <- 0

for (rep_trial in 1:itr) {
  
  xx <- runif(1)
  yy <- runif(1)
  if(yy <= xx*xx){
    xxx[rep_trial] <- xx
    yyy[rep_trial] <- yy
    counter <- counter + 1
  }
}
counter/itr

And the output (the integral, counter/itr) is 0.3312.

Simulations like these, the probability of hitting between the restricted boundaries of a unit square, are handy for estimating areas under functions that are difficult to integrate analytically.

At Duke University, two students had received A’s in chemistry all semester. But on the night before the final exam, they were partying in another state and didn’t get back to Duke until it was over. Their excuse to the professor was that they had a flat tire, and they asked if they could take a make-up test. The professor agreed, wrote out a test and sent the two to separate rooms to take it. The first question (on one side of the paper) was worth 5 points, and they answered it easily. Then they flipped the paper over and found the second question, worth 95 points: ‘Which tire was it?’

^{Source: Marilyn vos Savant, Parade Magazine, 3 March 1996, p 14.}

The question for today is: What is the probability for the students to get the second question right?
1) If the students have no preferences for a specific tyre?
2) If they have the following preference: right front, 58%, left front, 11%, right rear, 18%, left rear, 13%?

Counting

Let’s address the first problem by counting all possibilities.

There are a total of 16 possibilities, and four of them (highlighted in bold) are common. So the required probability for two students having the same random answers is 4/16 or 1/4.

Adding joint probabilities

Another way to reach the answer is by adding the joint probabilities of each of the four guesses. P(RF, RF) + P(LF, LF) + P(RR, RR) + P(LR, LR). P(RF, RF) is the joint probability of both the students arriving at the right front independently. If each one has a (1/4) chance, P(RF, RF) = P(RF) x P(RF) = (1/4)(1/4) = (1/16). Collecting all of them, P = P(RF) x P(RF) + P(LF) x P(LF) + P(RR) x P(RR) + P(LR) x P(LR) = (1/4)(1/4) + (1/4)(1/4) + (1/4)(1/4) + (1/4)*(1/4) = (1/4).

For the second question: P = 0.58 x 0.58 + 0.11 x 0.11 + 0.18 x 0.18 + 0.13 x 0.13 = 0.3978 or about 40%.

Simulation

Let’s run the following R code and verify the answer.

iter <- 10000
lie_count <- replicate(iter, {
   studentA <- sample(c(1,2,3,4), size = 1, replace = TRUE, prob = c(0.25, 0.25, 0.25, 0.25))
   studentB <- sample(c(1,2,3,4), size = 1, replace = TRUE, prob = c(0.25, 0.25, 0.25, 0.25))

   if(studentA == studentB){
     counter = 1
   }else{
     counter = 0
  }
})
mean(lie_count)

Output: 0.2532.

Change the probabilities, and you get the second one:

iter <- 10000
lie_count <- replicate(iter, {
   studentA <- sample(c(1,2,3,4), size = 1, replace = TRUE, prob = c(0.58, 0.11, 0.18, 0.13))
   studentB <- sample(c(1,2,3,4), size = 1, replace = TRUE, prob = c(0.58, 0.11, 0.18, 0.13))

   if(studentA == studentB){
      counter = 1
   }else{
      counter = 0
   }
})
mean(lie_count)

Output: 0.3971.