Lesson plan¶

2026-09-07 to and including 2026-09-11
Statistical Methods for Life Sciences

2026-08-31¶

This is when I teach:

Would you be able to join as a TA on Monday and Tuesday (Sept 7-8)? We mainly need TAs for the morning sessions on probability and distributions (Mon) and Linear regression (Tues). I hope you can join the course lunch (at Bikupan, BMC) as well!

Hence, that is the course materials

Probability Theory parts 1-4
Linear Regression sections 13-15

Both of these have exercises. I will just go through the exercises and see what happens :-) . I predict most problems will be because of R.

2026-09-01¶

At '1.3 Conditional probability' I could use an example for this:

Given E and F are subsets of S be two events with P(E) > 0, then the conditional probability of F given that E occurs is defined as...

I assume I will find it in the exercises. From Wikipedia

For example, the probability that any given person has a cough on any given day may be only 5%. But if we know or assume that the person is sick, then they are much more likely to be coughing. For example, the conditional probability that someone sick is coughing might be 75%, in which case we would have that P(Cough) = 5% and P(Cough|Sick) = 75%. Although there is a relationship between A and B in this example, such a relationship or dependence between A and B is not necessary, nor do they have to occur simultaneously.

E: coughing
F: being sick

So:

P(E) = 0.05 P(E | F) = 0.75 P(E | F) = P(E and F) / P(F) = 0.75 = unknown/unknown

Hmmm, maybe the examples have something I can solve :-)

probability mass function (PMF): the chance something happens, e.g. the chance to roll a 3.
cumulative distribution function (CDF): the chance to find a value or less than that value, e.g. the chance to roll a 3 or less

# Imagine a dice
expected_probabilities <- rep(1 / 6, 6)
found_probabilities <- seq(1, 6) / sum(seq(1, 6))

variance <- sum((expected_probabilities - found_probabilities) ^ 2) /
  length(expected_probabilities)
# 0.006613757

Blimey, this works:

Nheads <- replicate(10000, {
  coins <- sample(c("H", "T"), size=20, replace=TRUE)
  sum(coins == "H")
})
sum(Nheads >= 15)

replicate is a function name to remember :-)

While going through this, I miss a stated goal. Nor section 1, nor 2 starts with telling us which problem we are going to solve.

In the LOs I find:

understand the concept of random variables
understand the concept of probability
understand and learn to use resampling to compute probabilities
understand the concept probability mass function
understand the concept probability density function
understand the concept cumulative distribution functions
...

But nowhere I find what problems this solves.

The theory I fail to understand most is at section 2.3.3: I find it hard to even agree with the equations I see.

Exercises: Discrete random variables ¶

Exercise 1 (BRCA)¶

The probability of carrying one or more mutations in the breast cancer gene BRCA1 is 0.01. What is the probability of not carrying any mutations in BRCA1?

Yes and no are mutually exclusive, hence p_yes + p_no = 1.0. p_yes = 0.01 hence p_no = 1.0 - p_yes = 0.99

Exercise 2 (A coin toss)¶

When tossing a fair coin what is the probability of heads? what is the probability of tails?

0.5 and 0.5 P(H) = 0.5 P(T) = 0.5

Exercise 3 (Number of children)¶

In a region in Sweden with many children the number of children per household is between 0 and 6. The probability mass function is as follows;

x 0 1 2 3 4 5 6
p(x) 0.14 0.20 0.27 0.19 0.13 0.05 0.02

In a randomly chosen household

what is the probability of exactly 3 children?

P(X = 3) = 0.19

what is the probability of less than 3 children?

P(X < 3) = 0.14 + 0.20 + 0.27 = 0.61

what is the probability of 3 or less children?

P(X <= 3) = 0.14 + 0.20 + 0.27 + 0.19 = 0.8

what is the probability of an even number of children?

P(X % 2 == 0) = P(X == 0 || X == 2 || X == 4 || X == 6) = 0.14 + 0.27 + 0.13 + 0.02 = 0.56

Exercise 4 (Rolling dice)¶

When tossing a fair six-sided die

what is the probability of getting 6?

P(X = 6) = 1/6

what is the probability of an even number?

P(X is even) = 3/6

what is the probability of getting 3 or more?

P(X >= 3) = 4/6

what is the expected value of dots on the dice?

Q: Why P(X) and E[X], i.e. why E[X] has square brackets? Ah, this is just social. So, why did the teacher pick E[X] over E(X) ?

E[X] = mu = sum(of each element x in set S)(x * p(x)) where

x is the value of x
p(x) is the probability of finding x

E(X) = (1 1/6) + (2 1/6) + (3 1/6) + ... / 6 = 21 / 6 = 3.5

Exercise 5 (Randomization)¶

In a clinical trial, enrolled patients are randomly assigned to treatment or control group with equal probability.

For a single patient, what is the probability of being assigned to

the treatment group?

0.5

the control group?

0.5

If 20 patients are enrolled in the study; what is the probability of exactly 15 in the treatment group?

Need the probability mass function of the Bernouilli/Binomial distribution.

             (n)                    (20)
P(X = 15) =  (k) p^k (1-p)^(n -k) = (15) * 0.5^15 * 0.5^5 = (20! / (5! * 15!)) * 0.5^20 = 0.014785767

In R:

dbinom(x = 15, size = 20 , prob = 0.5) # 0.01478577

what is the probability of less than 7 in the treatment group?

# Sum the single probabilities
dbinom(x = 1, size = 20 , prob = 0.5) +
dbinom(x = 2, size = 20 , prob = 0.5) +
dbinom(x = 3, size = 20 , prob = 0.5) +
dbinom(x = 4, size = 20 , prob = 0.5) +
dbinom(x = 5, size = 20 , prob = 0.5) +
dbinom(x = 6, size = 20 , prob = 0.5) # 0.0576582

# Check that pbinom does that same. It does
pbinom(q = 6, size = 20, prob = 0.5) # 0.05765915

What is the most probable number of patients in the treatment group?

No equation for this, but n_trials p_success = 20 0.5 = 10

what is the probability of 5 or fewer patients in the control group?

pbinom(q = 5, size = 20, prob = 0.5) # 0.02069473

what is the probability of 2 or fewer patients in the treatment group?

pbinom(q = 2, size = 20, prob = 0.5) # 0.0002012253

Checking the answers, I think what is missing in the question:

Simulate 10000 clinical trials and count how many people are in the control group in each trail, so that you are able to conclude how often a trail will have 1 or 2 or 3 ... people in the control group

my answer:

set.seed(42)
n_trials <- 10000000
n_controls <- replicate(n = n_trials,
  sum(sample(x = c("C", "T"), size = 20, replace = TRUE) == "C")
)

Let's do the questions again