LESSON 3: MACHINE LEARNING Probability and Distributions

MACHINE LEARNING Probability and Distributions
Mastering a machine learning algorithm calls for understanding probability and distribution. Probabilities must be greater than or equal to zero. The sum of probabilities must be equal to one.
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[JosephRivera]

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[lawrencegayundato]

4:43 / 19:44
1)
X = {x1, x2, x3, ..., xn}
0 <= p(xi) <= 1
2)
p(x1) + p(x2) + p(x3) + ... + p(xn) = 1
btw, many thanks for your effort and time for making this great videos :) Nice one Sir!

[lawrencegayundato]

19:03 / 19:44
1) Costruct a Probability Distribution?
Solution:
Assuming that the random variable X is the sum of two face of fair dice.
X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
P(X=2) = P{(1, 1)}) = (1/6)(1/6) = 1/36
P(X=3) = P({(1, 2), (2, 1)}) = 2(1/6)(1/6) = 1/36 + 1/36 = 2/36
P(X=4) = P({(1, 3), (2, 2), (3, 1)}) = 3(1/6)(1/6) = 3/36
P(X=5) = P({(1, 4), (2, 3), (3, 2), (4, 1)}) = 4(1/6)(1/6) = 4/36
P(X=6) = P({(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}) = 5(1/6)(1/6) = 5/36
P(X=7) = P({(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}) = 7(1/6)(1/6) = 7/36
P(X=8) = P({(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}) = 5(1/6)(1/6) = 5/36
P(X=9) = P({(3, 6), (4, 5), (5, 4), (6, 3)}) = 4(1/6)(1/6) = 4/36
P(X=10) = P({(4, 6), (5, 5), (6, 4)}) = 3(1/6)(1/6) = 3/36
P(X=11) = P({(5, 6), (6, 5)}) = 2(1/6)(1/6) = 2/36
P(X=12) = P({(6, 6)}) = 1/36
2) Find P(X >= 11)?
Solution:
P(X >= 11) = P(X=11) + P(X=12) = 2/36 + 1/36 = 3/36 = 1/12 ≈ 0.0833
3) Find the probability that X takes an odd value.
Solution:

= 2/36 + 4/36 + 6/36 + 4/36 + 2/36
= (2 + 4 + 6 + 4 + 2)/36
= 18/36 = 0.5000 = 50%
Note: You don't need to compute for the distribution. Since the problem is if X takes an odd value. We let a new random variable Y such that X takes an odd or even value.
Y = {odd, even}
P(Y = odd) = 1/2
P(Y = even) = 1/2
P(Y = odd) + P(Y = even) = 1/2 + 1/2 = 1