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The sample space, \(\Omega\), is the collection of possible outcomes of an experiment, such as a die roll:
\[ \Omega = \{1, 2, 3, 4, 5, 6\} \]
An event, say \(E\), is a subset of \(\Omega\), such as the even dice rolls:
\[ E = \{2, 4, 6\} \]
An elementary or simple event is a particular result of an experiment, such as the roll of 4 and is represented as a lowercase omega.
\[ \omega = 4 \]
A null event or the empty set is represented as:
\[ \emptyset \].
\[ \omega \in E \]
implies that \(E\) occurs when \(\omega\) occurs; for example if \(\omega = 4\) occurs, then the event \(E\) (even number) occurs.
\[ \omega \not \in E \]
implies that \(E\) does not occur when \(\omega\) occurs; for example if \(\omega = 5\), \(E\) (even numbers) does not occur.
\[ E \subset F \]
implies that the occurrence of \(E\) implies the occurrence of \(F\); for example if \(E = \{2, 4, 5\}\) and \(F = \{2, 4, 5, 6\}\).
\[ E \cap F \]
implies the event that both \(E\) and \(F\) occur; for example if \(E = \{2, 4, 6\}\) (even numbers) and \(F = \{2, 3, 5\}\) (prime numbers), then \(E \cap F = 2\).
\[ E \cup F \]
implies the event that at least one of \(E\) or \(F\) occur; for example if \(E = \{2, 4, 6\}\) (even numbers) and \(F = \{2, 3, 5\}\) (prime numbers), then \(E \cup F = \{2, 3, 4, 5, 6\}\), which is that we get either an even number or a prime number or both.
\[ E \cap F = \emptyset \]
implies that \(E\) and \(F\) are mutually exclusive, or cannot both simultaneously occur; for example if \(E = \{2, 4, 6\}\) (even numbers) and \(F = \{1, 3, 5\}\) (odd numbers).
\[ E^\complement \]
is the event that \(E\) does not occur; for example if \(E\) are the even numbers then \(E^\complement\) are the odd numbers.
DeMorgan’s laws
\[ (A \cap B)^\complement = A^\complement \cup B^\complement\]
\[ (A \cup B)^\complement = A^\complement \cap B^\complement\]
To remember these laws, think of distributing the complements to \(A\) and \(B\) and then inverting the cap into a cup or vice versa.
\[ (A^\complement)^\complement = A \]
The complement of \(A\) complemented = \(A\).
And lastly:
\[ (A \cup B) \cap C = (A \cap C) \cup (B \cap C) \]
Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 17% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 12% while the probability that both the mother and father have contracted the disease is 6%. What is the probability that the mother has contracted influenza?
Let \(M\) be the mother and \(F\) be the father.
Recall that the union (\(\cup\)) implies the event that at least one of \(M\) or \(F\) occurred, therefore:
\[ P(M \cup F) = 17\%. \]
The probability that the father contracted influenza is 12%:
\[ P(F) = 12\%. \]
An intersection (\(\cap\)) implies the event that both M and F occur, i.e. the probability that both the mother and father have contracted the disease:
\[ P(M \cap F) = 6\%. \]
We know that:
\[ P(M \cup F) = P(M) + P(F) - P(M \cap F).\]
It is worth noting that if the events are mutually exclusive, \(P(M \cap F) = 0\), since they both can’t occur together, but in our case the events can occur together. From the above equation, we get:
\[ 17 = P(M) + 12 - 6 \\ P(M) = 17 - 12 + 6 \\ P(M) = 11%. \]
Therefore the probability of the mother contracting the disease is 11%.
You may be more familiar with the addition rule in probability, which is:
\[ P(M\ or\ F) = P(M) + P(F) - P(M\ and\ F) \\ 17 = P(M) + 12 - 6 \\ P(M) = 17 - 12 + 6 \\ P(M) = 11 \]
The probability of getting a 1 when rolling a fair die is \(\frac{1}{6}\). If we are given the extra information that the die roll was an odd number (1, 3 or 5), conditional on this new information, the probability of a 1 is now \(\frac{1}{3}\). Let’s test this example out:
Let \(B\) be an event so that \(P(B) > 0\), i.e. it’s an event that actually occurs. Then the conditional probability of an event \(A\) given that \(B\) has occurred is:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)}. \]
So using the example above:
\[ B = \{1, 3, 5\}\ and\ A = \{1\} \]
Therefore:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3} \]
sessionInfo()R version 4.3.0 (2023-04-21)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 22.04.2 LTS
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/openblas-pthread/libblas.so.3
LAPACK: /usr/lib/x86_64-linux-gnu/openblas-pthread/libopenblasp-r0.3.20.so; LAPACK version 3.10.0
locale:
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[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
time zone: Etc/UTC
tzcode source: system (glibc)
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] lubridate_1.9.2 forcats_1.0.0 stringr_1.5.0 dplyr_1.1.2
[5] purrr_1.0.1 readr_2.1.4 tidyr_1.3.0 tibble_3.2.1
[9] ggplot2_3.4.2 tidyverse_2.0.0 workflowr_1.7.0
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