Conditional Probabilities

Strings (SiPjAjk) = S7P6A64     Base Sequence = 12735     String Sequence = 12735 - 6 - 64

Expressions Of Pj Problems
Conditional Probabilities, Independent Events And Odds
Math A bag contains 12 balls (4 reds, 3 blues, 3 blacks, 2 greens) as indicated in Figure 118.4.
(a) Determine the probability of choosing a second blue ball from the bag given that the first ball chosen from the same bag is blue.
(b) Suppose there is a second bag containing 12 balls (4 reds, 3 blues, 3 blacks, 2 greens). Determine the probability of choosing a second blue ball from the second bag given that the first ball chosen from the first bag is blue.
(c) Determine the probability of chosing at least one blue ball, given both bags.
(d) What are the odds in favor of throwing a head in a single throw of a coin?
(e) What are the odds in favor of throwing at least one head on a single throw of two coins?

The strings: S7P6A64 (Grouping - Multi-criteria).

The math:
Pj Problem of Interest is of type grouping (multi-criteria). Grouping is at the heart of statistics. The grouping may be permutational or combinational, single criterion or multi-criteria grouping. Formulas of Interest
Consider two independent events A and B with probabilities of occurrence P[A] and P[B] respectively.
Probability of both A and B occurring is P[A∩B]:
P[A∩B] = P[AB] = P[A] x P[B] -------(1)
Probability of either A or B occuring is P[A∪B]:
P[A∪B] = P[A+B] = P[A] + P[B] - P[A∩B] --------(2)

If the occurrence of A is dependent on the occurrence of B
Conditional Probability of B occurring given that A has occurred is P[B|A]:
P[B|A] = P[AB]/P[A] = (P[A] x P[B])/P[A] -------(3)

Odds means the ratio of the probability in favor of an event to the probability against the event.

(a) Let P[A] = probability first ball chosen from the bag is blue
Then, P[A] = 3/12 = 1/4
Let probability of choosing a second blue ball from the bag given that the first ball chosen from the same bag is blue be P[B|A]
Then conditional probabily, P[B|A] = 2/12 = 1/6.

(b) Let event G = first ball from first bag is blue
So, P[G] = 3/12
Let event H = second ball from second bag is blue
So, P[H] = 3/12
G and H are independent
So, P[GH] = (3/12) x (3/12) = 1/16.

(c) Probability of least one blue ball is P[A∪B]:
P[A∪B]= P[A+B] = P[A] + P[B] - P[A∩B] = 1/4 + 1/4 - 1/16 = 7/16.

(d) Possibilities: H, T. Where H is head and T is tail.
So, probability in favor of a head in one throw of a coin = 1/2
Probability against = 1/2
So odds = (1/2)/(1/2) = 1.

(e) Possibilities: TT, TH, HT, HH
So, probability in favor of at least one head in a single throw of two coins = 3/4
Probability against = 1/4
So, odds = (3/4)/(1/4) = 3/1 = 3:1. The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
Single Variable Functions
Conics
Ordinary Differential Equations (ODEs)
Vector Spaces
Real Numbers The Universe is composed of matter and radiant energy. Matter is any kind of mass-energy that moves with velocities less than the velocity of light. Radiant energy is any kind of mass-energy that moves with the velocity of light.
Periodic Table
Composition And Structure Of Matter
How Matter Gets Composed
How Matter Gets Composed (2)
Molecular Structure Of Matter
Molecular Shapes: Bond Length, Bond Angle
Molecular Shapes: Valence Shell Electron Pair Repulsion
Molecular Shapes: Orbital Hybridization
Molecular Shapes: Sigma Bonds Pi Bonds
Molecular Shapes: Non ABn Molecules
Molecular Orbital Theory
More Pj Problem Strings

Blessed are they that have not seen, and yet have believed. John 20:29

TECTechnic Logo, Kimberlee J. Benart | © 2018 | All rights reserved | Founder and Site Programmer, Peter O. Sagay.