Linear Algebra
&
Engineering Mathematics 1
Week 10
Remarks on Probability & Linear Algebra
Remarks from workshop problems
Problem 1: A railway goods shed expects trains to arrive on time 80% of the time and depart on time 75% of the time. If the chance of a train departing on time is 85% when it arrived on time, what must be the chance of it departing on time when it arrived late?
Remarks from workshop problems
Problem 1: A railway goods shed expects trains to arrive on time 80% of the time and depart on time 75% of the time. If the chance of a train departing on time is 85% when it arrived on time, what must be the chance of it departing on time when it arrived late?
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$A = $ arrives on time. $D = $ departs on time. We have: |
Can we use directly the definition of conditional probability?
$P(D\,|\, A')= \dfrac{P(D \text{ and } A')}{P(A')}$ $\qquad\quad\;\;\,=\dfrac{P(D\,| \,A')P(A')}{P(A')}$ ❌ No, we can't! ❌ |
Remarks from workshop problems
Problem 1: A railway goods shed expects trains to arrive on time 80% of the time and depart on time 75% of the time. If the chance of a train departing on time is 85% when it arrived on time, what must be the chance of it departing on time when it arrived late?
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$P(A) = 0.8$ |
In this case we must use:
$P(D) = P(D\,|\,A)P(A) + P(D\,|\,A')P(A')$ 
Tree diagram 👆 |
Remarks from workshop problems
Problem 2: What is the chance of at least 2 people out of 6 sharing a birth hour? (you may assume births are equally distributed throughout the day).
Let's calculate the number of ways to assign distinct hours to 6 people: \[24\times 23 \times 22 \times 21 \times 20\times 19\]
The total number of possible arrangements (allowing repeated hours) is: \[24^6\]
So the probability that all 6 have different birth hours is: \[ \small P(\text{all different})= \dfrac{24\times 23 \times 22 \times 21 \times 20\times 19}{24^6}\approx 0.5073 \]
Remarks from workshop problems
Problem 2: What is the chance of at least 2 people out of 6 sharing a birth hour? (you may assume births are equally distributed throughout the day).
So the probability that all 6 have different birth hours is: \[ \small P(\text{all different})= \dfrac{24\times 23 \times 22 \times 21 \times 20\times 19}{24^6}\approx 0.5073 \]
Thus \[ \small P(\text{at least 2 share a birth hour})= 1- 0.5073 = 0.4927 =49.27\% \]
Remarks from workshop problems
Problem 2: What is the chance of at least 2 people out of 6 sharing a birth hour? (you may assume births are equally distributed throughout the day).
\[ P(\text{at least 2 share a birth hour})= 1- P(\text{all different})= 49.27\% \]
1. Why use permutations, not combinations?
Ans. Each person is an individual, distinguishable entity (e.g., Person 1, Person 2, ..., Person 6). We're considering who gets which hour.
2. What is the probability of at least 3 people out of 6 sharing a birth hour?
Ans. Here we can use the same the complement method:
$P(\text{at least 3 share...})$ $\,= 1\, - $ $ P(\text{at most 2 share any hour}).$
This is actually harder to answer. 🤯
Determinants
The determinant is number assigned to square matrices that measures how the corresponding linear mapping stretches the space. In particular, this number, can be used to test for invertibility of a matrix.
Linear transformations in 2D
Linear transformations in 3D
Source: Matrix transformations
Affine transformations
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$ \ds f(x,y)=\begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} e \\ f \end{bmatrix} $ $ \ds f_1(x,y)=\begin{bmatrix} 0.00 & 0.00 \\ 0.00 & 0.16 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $ $ \ds f_2(x,y)=\begin{bmatrix} 0.85 & 0.04 \\ -0.04 & 0.85 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0.00 \\ 1.60 \end{bmatrix} $ $ \ds f_3(x,y)=\begin{bmatrix} 0.20 & -0.26 \\ 0.23 & 0.22 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0.00 \\ 1.60 \end{bmatrix} $ $ \ds f_4(x,y)=\begin{bmatrix} -0.15 & 0.28 \\ 0.26 & 0.24 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0.00 \\ 0.44 \end{bmatrix} $ |
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Barnsley fern in JavaScript
Click on Play. Explore: Modify the code and click again on
Play to see the changes. Have fun! 🌿 🤓
AI - Machine Learning
| Linear Algebra \(A \mathbf x = \mathbf b\) |
Probability \(P(B\,|\, A)\) |
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| Calculus \(\nabla f(x,y)\) |
Statistics 📊📈 |
Practice Gauss elimination
Solve the system and input your solution for $x,$ $y,$ and $z.$ Then, click on Check. Source