Chapter 3 Conditional probability

This chapter deals with conditional probability.

The students are expected to acquire the following knowledge:

Theoretical

Identify whether variables are independent.
Calculation of conditional probabilities.
Understanding of conditional dependence and independence.
How to apply Bayes’ theorem to solve difficult probabilistic questions.

Simulating conditional probabilities.
cumsum.
apply.

3.1 Calculating conditional probabilities

Exercise 3.1 A military officer is in charge of identifying enemy aircraft and shooting them down. He is able to positively identify an enemy airplane 95% of the time and positively identify a friendly airplane 90% of the time. Furthermore, 99% of the airplanes are friendly. When the officer identifies an airplane as an enemy airplane, what is the probability that it is not and they will shoot at a friendly airplane?

Solution. Let $E = 0$ denote that the observed plane is friendly and $E=1$ that it is an enemy. Let $I = 0$ denote that the officer identified it as friendly and $I = 1$ as enemy. Then

\[\begin{align} P(E = 0 | I = 1) &= \frac{P(I = 1 | E = 0)P(E = 0)}{P(I = 1)} \\ &= \frac{P(I = 1 | E = 0)P(E = 0)}{P(I = 1 | E = 0)P(E = 0) + P(I = 1 | E = 1)P(E = 1)} \\ &= \frac{0.1 \times 0.99}{0.1 \times 0.99 + 0.95 \times 0.01} \\ &= 0.91. \end{align}\]

Exercise 3.2 R: Consider tossing a fair die. Let $A = \{2,4,6\}$ and $B = \{1,2,3,4\}$. Then $P(A) = \frac{1}{2}$, $P(B) = \frac{2}{3}$ and $P(AB) = \frac{1}{3}$. Since $P(AB) = P(A)P(B)$, the events $A$ and $B$ are independent. Simulate draws from the sample space and verify that the proportions are the same. Then find two events $C$ and $D$ that are not independent and repeat the simulation.

set.seed(1)
nsamps <- 10000
tosses <- sample(1:6, nsamps, replace = TRUE)
PA <- sum(tosses %in% c(2,4,6)) / nsamps
PB <- sum(tosses %in% c(1,2,3,4)) / nsamps
PA * PB

## [1] 0.3295095

sum(tosses %in% c(2,4)) / nsamps

## [1] 0.3323

# Let C = {1,2} and D = {2,3}
PC <- sum(tosses %in% c(1,2)) / nsamps
PD <- sum(tosses %in% c(2,3)) / nsamps
PC * PD

## [1] 0.1067492

sum(tosses %in% c(2)) / nsamps

## [1] 0.1622

Exercise 3.3 A machine reports the true value of a thrown 12-sided die 5 out of 6 times.

If the machine reports a 1 has been tossed, what is the probability that it is actually a 1?
Now let the machine only report whether a 1 has been tossed or not. Does the probability change?
R: Use simulation to check your answers to a) and b).

Solution.

Let $T = 1$ denote that the toss is 1 and $M = 1$ that the machine reports a 1. \[\begin{align} P(T = 1 | M = 1) &= \frac{P(M = 1 | T = 1)P(T = 1)}{P(M = 1)} \\ &= \frac{P(M = 1 | T = 1)P(T = 1)}{\sum_{k=1}^{12} P(M = 1 | T = k)P(T = k)} \\ &= \frac{\frac{5}{6}\frac{1}{12}}{\frac{5}{6}\frac{1}{12} + 11 \frac{1}{6} \frac{1}{11} \frac{1}{12}} \\ &= \frac{5}{6}. \end{align}\]
Yes. \[\begin{align} P(T = 1 | M = 1) &= \frac{P(M = 1 | T = 1)P(T = 1)}{P(M = 1)} \\ &= \frac{P(M = 1 | T = 1)P(T = 1)}{\sum_{k=1}^{12} P(M = 1 | T = k)P(T = k)} \\ &= \frac{\frac{5}{6}\frac{1}{12}}{\frac{5}{6}\frac{1}{12} + 11 \frac{1}{6} \frac{1}{12}} \\ &= \frac{5}{16}. \end{align}\]

set.seed(1)
nsamps   <- 10000
report_a <- vector(mode = "numeric", length = nsamps)
report_b <- vector(mode = "logical", length = nsamps)
truths   <- vector(mode = "logical", length = nsamps)
for (i in 1:10000) {
  toss      <- sample(1:12, size = 1)
  truth     <- sample(c(TRUE, FALSE), size = 1, prob = c(5/6, 1/6))
  truths[i] <- truth
  if (truth) {
    report_a[i] <- toss
    report_b[i] <- toss == 1
  } else {
    remaining   <- (1:12)[1:12 != toss]
    report_a[i] <- sample(remaining, size = 1)
    report_b[i] <- toss != 1
  }
}
truth_a1 <- truths[report_a == 1]
sum(truth_a1) / length(truth_a1)

## [1] 0.8300733

truth_b1 <- truths[report_b]
sum(truth_b1) / length(truth_b1)

## [1] 0.3046209

Exercise 3.4 A coin is tossed independently $n$ times. The probability of heads at each toss is $p$. At each time $k$, $(k = 2,3,...,n)$ we get a reward at time $k+1$ if $k$-th toss was a head and the previous toss was a tail. Let $A_k$ be the event that a reward is obtained at time $k$.

Are events $A_k$ and $A_{k+1}$ independent?
Are events $A_k$ and $A_{k+2}$ independent?
R: simulate 10 tosses 10000 times, where $p = 0.7$. Check your answers to a) and b) by counting the frequencies of the events $A_5$, $A_6$, and $A_7$.

Solution.

For $A_k$ to happen, we need the tosses $k-2$ and $k-1$ be tails and heads respectively. For $A_{k+1}$ to happen, we need tosses $k-1$ and $k$ be tails and heads respectively. As the toss $k-1$ need to be heads for one and tails for the other, these two events can not happen simultaneously. Therefore the probability of their intersection is 0. But the probability of each of them separately is $p(1-p) > 0$. Therefore, they are not independent.
For $A_k$ to happen, we need the tosses $k-2$ and $k-1$ be tails and heads respectively. For $A_{k+2}$ to happen, we need tosses $k$ and $k+1$ be tails and heads respectively. So the probability of intersection is $p^2(1-p)^2$. And the probability of each separately is again $p(1-p)$. Therefore, they are independent.

set.seed(1)
nsamps <- 10000
p      <- 0.7
rewardA_5  <- vector(mode = "logical", length = nsamps)
rewardA_6  <- vector(mode = "logical", length = nsamps)
rewardA_7  <- vector(mode = "logical", length = nsamps)
rewardA_56 <- vector(mode = "logical", length = nsamps)
rewardA_57 <- vector(mode = "logical", length = nsamps)
for (i in 1:nsamps) {
  samps <- sample(c(0,1), size = 10, replace = TRUE, prob = c(0.7, 0.3))
  rewardA_5[i]  <- (samps[4] == 0 & samps[3] == 1)
  rewardA_6[i]  <- (samps[5] == 0 & samps[4] == 1)
  rewardA_7[i]  <- (samps[6] == 0 & samps[5] == 1)
  rewardA_56[i] <- (rewardA_5[i] & rewardA_6[i])
  rewardA_57[i] <- (rewardA_5[i] & rewardA_7[i])
}
sum(rewardA_5) / nsamps

## [1] 0.2141

sum(rewardA_6) / nsamps

## [1] 0.2122

sum(rewardA_7) / nsamps

## [1] 0.2107

sum(rewardA_56) / nsamps

## [1] 0

sum(rewardA_57) / nsamps

## [1] 0.0454

Exercise 3.5 A drawer contains two coins. One is an unbiased coin, the other is a biased coin, which will turn up heads with probability $p$ and tails with probability $1-p$. One coin is selected uniformly at random.

The selected coin is tossed $n$ times. The coin turns up heads $k$ times and tails $n-k$ times. What is the probability that the coin is biased?
The selected coin is tossed repeatedly until it turns up heads $k$ times. Given that it is tossed $n$ times in total, what is the probability that the coin is biased?

Solution.

Let $B = 1$ denote that the coin is biased and let $H = k$ denote that we’ve seen $k$ heads. \[\begin{align} P(B = 1 | H = k) &= \frac{P(H = k | B = 1)P(B = 1)}{P(H = k)} \\ &= \frac{P(H = k | B = 1)P(B = 1)}{P(H = k | B = 1)P(B = 1) + P(H = k | B = 0)P(B = 0)} \\ &= \frac{p^k(1-p)^{n-k} 0.5}{p^k(1-p)^{n-k} 0.5 + 0.5^{n+1}} \\ &= \frac{p^k(1-p)^{n-k}}{p^k(1-p)^{n-k} + 0.5^n}. \end{align}\]
The same results as in a). The only difference between these two scenarios is that in b) the last throw must be heads. However, this holds for the biased and the unbiased coin and therefore does not affect the probability of the coin being biased.

Exercise 3.6 Judy goes around the company for Women’s day and shares flowers. In every office she leaves a flower, if there is at least one woman inside. The probability that there’s a woman in the office is $\frac{3}{5}$.

What is the probability that Judy leaves her first flower in the fourth office?
Given that she has given away exactly three flowers in the first four offices, what is the probability that she gives her fourth flower in the eighth office?
What is the probability that she leaves the second flower in the fifth office?
What is the probability that she leaves the second flower in the fifth office, given that she did not leave the second flower in the second office?
Judy needs a new supply of flowers immediately after the office, where she gives away her last flower. What is the probability that she visits at least five offices, if she starts with two flowers?
R: simulate Judy’s walk 10000 times to check your answers a) - e).

Solution. Let $X_i = k$ denote the event that … $i$-th sample on the $k$-th run.

Since the events are independent, we can multiply their probabilities to get \[\begin{equation} P(X_1 = 4) = 0.4^3 \times 0.6 = 0.0384. \end{equation}\]
Same as in a) as we have a fresh start after first four offices.
For this to be possible, she had to leave the first flower in one of the first four offices. Therefore there are four possibilities, and for each of those the probability is $0.4^3 \times 0.6$. Additionally, the probability that she leaves a flower in the fifth office is $0.6$. So \[\begin{equation} P(X_2 = 5) = \binom{4}{1} \times 0.4^3 \times 0.6^2 = 0.09216. \end{equation}\]
We use Bayes’ theorem. \[\begin{align} P(X_2 = 5 | X_2 \neq 2) &= \frac{P(X_2 \neq 2 | X_2 = 5)P(X_2 = 5)}{P(X_2 \neq 2)} \\ &= \frac{0.09216}{0.64} \\ &= 0.144. \end{align}\] The denominator in the second equation can be calculated as follows. One of three things has to happen for the second not to be dealt in the second round. First, both are zero, so $0.4^2$. Second, first is zero, and second is one, so $0.4 \times 0.6$. Third, the first is one and the second one zero, so $0.6 \times 0.4$. Summing these values we get $0.64$.
We will look at the complement, so the events that she gave away exactly two flowers after two, three and four offices. \[\begin{equation} P(X_2 \geq 5) = 1 - 0.6^2 - 2 \times 0.4 \times 0.6^2 - 3 \times 0.4^2 \times 0.6^2 = 0.1792. \end{equation}\] The multiplying parts represent the possibilities of the first flower.

set.seed(1)
nsamps     <- 100000
Judyswalks <- matrix(data = NA, nrow = nsamps, ncol = 8)
for (i in 1:nsamps) {
  thiswalk <- sample(c(0,1), size = 8, replace = TRUE, prob = c(0.4, 0.6))
  Judyswalks[i, ] <- thiswalk
}
csJudy <- t(apply(Judyswalks, 1, cumsum))

# a
sum(csJudy[ ,4] == 1 & csJudy[ ,3] == 0) / nsamps

## [1] 0.03848

# b
csJsubset <- csJudy[csJudy[ ,4] == 3 & csJudy[ ,3] == 2, ]
sum(csJsubset[ ,8] == 4 & csJsubset[ ,7] == 3) / nrow(csJsubset)

## [1] 0.03665893

# c
sum(csJudy[ ,5] == 2 & csJudy[ ,4] == 1) / nsamps

## [1] 0.09117

# d
sum(csJudy[ ,5] == 2 & csJudy[ ,4] == 1) / sum(csJudy[ ,2] != 2)

## [1] 0.1422398

# e
sum(csJudy[ ,4] < 2) / nsamps

## [1] 0.17818

3.2 Conditional independence

Exercise 3.7 Describe:

A real-world example of two events $A$ and $B$ that are dependent but become conditionally independent if conditioned on a third event $C$.
A real-world example of two events $A$ and $B$ that are independent, but become dependent if conditioned on some third event $C$.

Solution.

Let $A$ be the height of a person and let $B$ be the person’s knowledge of the Dutch language. These events are dependent since the Dutch are known to be taller than average. However if $C$ is the nationality of the person, then $A$ and $B$ are independent given $C$.
Let $A$ be the event that Mary passes the exam and let $B$ be the event that John passes the exam. These events are independent. However, if the event $C$ is that Mary and John studied together, then $A$ and $B$ are conditionally dependent given $C$.

Exercise 3.8 We have two coins of identical appearance. We know that one is a fair coin and the other flips heads 80% of the time. We choose one of the two coins uniformly at random. We discard the coin that was not chosen. We now flip the chosen coin independently 10 times, producing a sequence $Y_1 = y_1$, $Y_2 = y_2$, …, $Y_{10} = y_{10}$.

Intuitively, without doing and computation, are these random variables independent?
Compute the probability $P(Y_1 = 1)$.
Compute the probabilities $P(Y_2 = 1 | Y_1 = 1)$ and $P(Y_{10} = 1 | Y_1 = 1,...,Y_9 = 1)$.
Given your answers to b) and c), would you now change your answer to a)? If so, discuss why your intuition had failed.

Solution.

$P(Y_1 = 1) = 0.5 * 0.8 + 0.5 * 0.5 = 0.65$.
Since we know that $Y_1 = 1$ this should change our view of the probability of the coin being biased or not. Let $B = 1$ denote the event that the coin is biased and let $B = 0$ denote that the coin is unbiased. By using marginal probability, we can write \[\begin{align} P(Y_2 = 1 | Y_1 = 1) &= P(Y_2 = 1, B = 1 | Y_1 = 1) + P(Y_2 = 1, B = 0 | Y_1 = 1) \\ &= \sum_{k=1}^2 P(Y_2 = 1 | B = k, Y_1 = 1)P(B = k | Y_1 = 1) \\ &= 0.8 \frac{P(Y_1 = 1 | B = 1)P(B = 1)}{P(Y_1 = 1)} + 0.5 \frac{P(Y_1 = 1 | B = 0)P(B = 0)}{P(Y_1 = 1)} \\ &= 0.8 \frac{0.8 \times 0.5}{0.65} + 0.5 \frac{0.5 \times 0.5}{0.65} \\ &\approx 0.68. \end{align}\] For the other calculation we follow the same procedure. Let $X = 1$ denote that first nine tosses are all heads (equivalent to $Y_1 = 1$,…, $Y_9 = 1$). \[\begin{align} P(Y_{10} = 1 | X = 1) &= P(Y_2 = 1, B = 1 | X = 1) + P(Y_2 = 1, B = 0 | X = 1) \\ &= \sum_{k=1}^2 P(Y_2 = 1 | B = k, X = 1)P(B = k | X = 1) \\ &= 0.8 \frac{P(X = 1 | B = 1)P(B = 1)}{P(X = 1)} + 0.5 \frac{P(X = 1 | B = 0)P(B = 0)}{P(X = 1)} \\ &= 0.8 \frac{0.8^9 \times 0.5}{0.5 \times 0.8^9 + 0.5 \times 0.5^9} + 0.5 \frac{0.5^9 \times 0.5}{0.5 \times 0.8^9 + 0.5 \times 0.5^9} \\ &\approx 0.8. \end{align}\]

3.3 Monty Hall problem

The Monty Hall problem is a famous probability puzzle with non-intuitive outcome. Many established mathematicians and statisticians had problems solving it and many even disregarded the correct solution until they’ve seen the proof by simulation. Here we will show how it can be solved relatively simply with the use of Bayes’ theorem if we select the variables in a smart way.

Exercise 3.9 (Monty Hall problem) A prize is placed at random behind one of three doors. You pick a door. Now Monty Hall chooses one of the other two doors, opens it and shows you that it is empty. He then gives you the opportunity to keep your door or switch to the other unopened door. Should you stay or switch? Use Bayes’ theorem to calculate the probability of winning if you switch and if you do not. R: Check your answers in R.

Solution. W.L.O.G. assume we always pick the first door. The host can only open door 2 or door 3, as he can not open the door we picked. Let $k \in \{2,3\}$.

Let us first look at what happens if we do not change. Then we have

\[\begin{align} P(\text{car in 1} | \text{open $k$}) &= \frac{P(\text{open $k$} | \text{car in 1})P(\text{car in 1})}{P(\text{open $k$})} \\ &= \frac{P(\text{open $k$} | \text{car in 1})P(\text{car in 1})}{\sum_{n=1}^3 P(\text{open $k$} | \text{car in $n$})P(\text{car in $n$)}}. \end{align}\]

The probability that he opened $k$ if the car is in 1 is $\frac{1}{2}$, as he can choose between door 2 and 3 as both have a goat behind it. Let us look at the normalization constant. When $n = 1$ we get the value in the nominator. When $n=k$, we get 0, as he will not open the door if there’s a prize behind. The remaining option is that we select 1, the car is behind $k$ and he opens the only door left. Since he can’t open 1 due to it being our pick and $k$ due to having the prize, the probability of opening the remaining door is 1, and the prior probability of the car being behind this door is $\frac{1}{3}$. So we have

\[\begin{align} P(\text{car in 1} | \text{open $k$}) &= \frac{\frac{1}{2}\frac{1}{3}}{\frac{1}{2}\frac{1}{3} + \frac{1}{3}} \\ &= \frac{1}{3}. \end{align}\]

Now let us look at what happens if we do change. Let $k' \in \{2,3\}$ be the door that is not opened. If we change, we select this door, so we have

\[\begin{align} P(\text{car in $k'$} | \text{open $k$}) &= \frac{P(\text{open $k$} | \text{car in $k'$})P(\text{car in $k'$})}{P(\text{open $k$})} \\ &= \frac{P(\text{open $k$} | \text{car in $k'$})P(\text{car in $k'$})}{\sum_{n=1}^3 P(\text{open $k$} | \text{car in $n$})P(\text{car in $n$)}}. \end{align}\]

The denominator stays the same, the only thing that is different from before is $P(\text{open $k$} | \text{car in $k'$})$. We have a situation where we initially selected door 1 and the car is in door $k'$. The probability that the host will open door $k$ is then 1, as he can not pick any other door. So we have

\[\begin{align} P(\text{car in $k'$} | \text{open $k$}) &= \frac{\frac{1}{3}}{\frac{1}{2}\frac{1}{3} + \frac{1}{3}} \\ &= \frac{2}{3}. \end{align}\]

Therefore it makes sense to change the door.

set.seed(1)
nsamps <- 1000
ifchange <- vector(mode = "logical", length = nsamps)
ifstay   <- vector(mode = "logical", length = nsamps)
for (i in 1:nsamps) {
  where_car      <- sample(c(1:3), 1)
  where_player   <- sample(c(1:3), 1)
  open_samp      <- (1:3)[where_car != (1:3) & where_player != (1:3)]
  if (length(open_samp) == 1) {
    where_open <- open_samp
  } else {
      where_open <- sample(open_samp, 1)
  }
  ifstay[i]      <- where_car == where_player
  where_ifchange <- (1:3)[where_open != (1:3) & where_player != (1:3)]
  ifchange[i]    <- where_ifchange == where_car
}
sum(ifstay) / nsamps

## [1] 0.328

sum(ifchange) / nsamps

## [1] 0.672