Probability - d2l.ai Exercises - Part 6
This notebook tries to solve the probability exercises in d2l.ai
- Prepare libraries, variables and functions for plotting
- Vary the number of samples and number of groups for the dice experiment. Observe and analyze the experimental results.
- Given two events with probability $P(A)$ and $P(B)$ , compute upper and lower bounds on $P(A∪B)$ and $P(A∩B)$ .
- Assume that we have a sequence of random variables, say A , B , and C , where B only depends on A , and C only depends on B , can you simplify the joint probability P(A,B,C) ?
- In Section 2.6.2.6, the first test is more accurate. Why not run the first test twice rather than run both the first and second tests?
%matplotlib inline
import numpy as np
import tensorflow as tf
import tensorflow_probability as tfp
from d2l import tensorflow as d2l
from matplotlib_venn import venn2
def plot_estimates(estimates, title_str):
d2l.set_figsize((6, 4.5))
for i in range(6):
d2l.plt.plot(estimates[:, i].numpy(), label=("P(die=" + str(i + 1) + ")"))
d2l.plt.axhline(y=0.167, color='black', linestyle='dashed')
d2l.plt.gca().set_xlabel('Groups of experiments')
d2l.plt.gca().set_ylabel('Estimated probability')
d2l.plt.legend();
d2l.plt.title(title_str)
fair_probs = tf.ones(6) / 6
From this we can see that the more number of samples/experiment groups that we perform, the probability that a number will fall on the die getting close to 1/6.
Let's sample 1000 from two events with $P(A) =0.5$ and $P(B) =0.7$ then represent the results in a venn diagram as hinted.
counts = tfp.distributions.Binomial(1, probs=[.5, .7]).sample(1000)
event_samples = list(counts.numpy().astype(int).tolist())
event_10 = event_samples.count([1,0])
event_01 = event_samples.count([0,1])
event_11 = event_samples.count([1,1])
venn2((event_10, event_01, event_11), ('Event 1', 'event 2'))
d2l.plt.show()
From the diagram we can see that $$ P(A) = 468/1000 = 0.468 (\approx 0.5) \\ P(B) = 692/1000 = 0.692 (\approx 0.7)$$
Assuming Upper bound for $P(A∪B) = 1$, so that $$P(A∩B) >= P(A) + P(B) - 1 \\P(A∩B) >= 0.16$$ so the lower bound of $P(A∩B)$ is 0.16.
The upper bound of $P(A∩B)$ would be $min(P(A), P(B))$, since the $P(A∩B)$ is event of both A and B ocurring together. Consequently lower bound of $P(A∪B)$ would be $max(P(A), P(B))$
so, $$P(A∩B) <= 0.468$$ and $$P(A∪B) >= 0.692$$
At last $$0.692 <= P(A∪B) <= 1$$ and $$0.16<=P(A∩B)<=0.468$$
Since this is a sequential occurence, A occurred first, B occurred second and at last C occurred. Let's assume $A, B$ as one event $Z$
$$P(Z, C) = P(C|Z) * P(Z) $$
substituting value for $Z$ $$P(A, B, C) = P(C|A, B) * P(A, B) $$ $$P(A, B, C) = P(C|A, B) * P(A|B) * P(B) $$
In Section 2.6.2.6, the first test is more accurate. Why not run the first test twice rather than run both the first and second tests?
We did not conduct the same test because simply put running the same test on the same person, will not give a different result.