Understanding Probability

 A Basic Introduction

Likelihood is a crucial idea in science and measurements that actions the probability of an occasion happening. It evaluates vulnerability and is utilized widely in different fields, including science, designing, finance, and regular direction.

Basic Definitions

1. Experiment: An action or process that generates outcomes. For example, tossing a coin or rolling a die.
2. Sample Space (S): The set of all possible outcomes of an experiment. For a coin toss, the sample space is {Heads, Tails}.
3. Event (E): A subset of the sample space. An event might consist of one or more outcomes. For example, getting a Head in a coin toss.

Probability of an Event

The likelihood of an occasion

E is characterized as the proportion of the quantity of ideal results to the all out number of potential results in the example space. It is determined utilizing the recipe:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

For a fair six-sided die, the probability of rolling a 3 is:

$P(\text{Rolling a 3}) = \frac{1}{6}$

Properties of Probability

1. Non-negativity: Probability values are always between 0 and 1.

   • $0 \leq P(E) \leq 1$
   • $P(E) = 0$ implies the event will not occur.
   • $P(E) = 1$ implies the event will certainly occur.

2. Sum of Probabilities: The sum of probabilities of all mutually exclusive events in a sample space is 1.

   • For a single coin toss: $P(\text{Heads}) + P(\text{Tails}) = 1$

3. Complementary Rule: The probability of the complement of an event $E$ (denoted as $E'$) is:

   • $P(E') = 1 - P(E)$

Types of Events

1. Independent Events: Two events are independent if the occurrence of one does not affect the occurrence of the other. For example, tossing two separate coins.

   • $P(A \cap B) = P(A) \cdot P(B)$

2. Dependent Events: Two events are dependent if the occurrence of one affects the occurrence of the other. For example, drawing cards from a deck without replacement.

3. Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur simultaneously. For example, rolling a die and getting both an even and an odd number at the same time.

   • $P(A \cup B) = P(A) + P(B)$

Conditional Probability

The probability of an event $A$ given that another event $B$ has occurred is called conditional probability and is denoted as $P(A|B)$. It is calculated using:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

Bayes’ Theorem

Bayes' Theorem relates the conditional and marginal probabilities of random events. It is a powerful tool for calculating the probability of an event based on prior knowledge of conditions related to the event.

$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$

Conclusion

Likelihood is a urgent idea that helps us comprehend and evaluate vulnerability. It gives a numerical system to going with forecasts and educated choices in light of the probability regarding different results. Whether applied in logical examination, risk evaluation, or regular situations, understanding likelihood upgrades our capacity to explore a questionable world.

Example

1: Show that if E and F are independent events, then E’ and F’ are independent events.

Ans:

Step 1

DEFINITIONS

Two events E1 and  E1 are independent independent if and only if:

P(E1∩E2)=P(E1)⋅P(E2)

Complement rule Complement rule:

P(E)=1−P(E)

General addition rule General addition rule for any two events:

P(A∪B)=P(A)+P(B)−P(A∩B)

Step 2

Given: E and F are independent

To proof: ‾E and ‾F are independent

PROOFPROOF

Since E and F are independent:

P(E∩F)=P(E)⋅P(F)

Use the complement rule:

P( ‾E)=1−P(E)

P( ‾F)=1−P(F)

Let us determine their product:

P(E)⋅P(F)=(1−P(E))(1−P(F))    

=1−P(E)−P(F)+P(E)⋅P(F)      Using Distributive property

=1−P(E)−P(F)+PS(E∩F)        Since E and F are independent event  

1−(P(E)+P(F)−P(E∩F))

=1−P(E∪F)      General addition rule

=P(E∪F)           Complement rule

=P(E∩F)         EUF=EUF

    Since P(E)⋅P(F)= P(E∩F), E and F are independent

2: What is the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up heads?

Ans

DEFINITIONS

Product rule Product rule If one event can occur in m ways AND a second event can occur in n ways, then the number of ways that the two events can occur in sequence is then m⋅n.

Definition Conditional probability:

P(B∣A))=(p(AUB))/(P(A))

Definition permutation permutation (order is important):

P(n,r)=n!/(n-r)!

Definition combination combination (order is not important)

C(n,r)=n!/r!(n-r)!

with n!=n⋅(n−1)⋅...⋅2⋅1

Step 2

 

Result

                                                    

3: What is the expected value when a $1 lottery ticket is bought in which the purchaser wins exactly $10 million if the ticket contains the six winning numbers chosen from the set {1,2,3,....,50} and the purchaser wins nothing otherwise?

Ans:

DEFINITIONS

Definition permutation permutation (order is important):

P(n,r)=n!/(n-r)!

Definition combination combination (order is not important):

C(n,r)=n!/r!(n-r)!

With n!=n⋅(n−1)⋅...⋅2⋅1

Complement rule:

P(E)=1−P(E)

SOLUTION

The order of the numbers is not important (since a different order results in the same numbers being picked) and thus we need to use the definition of combination combination.

We need to select 6 numbers from the 50 numbers {1,2,3,...,50}{1,2,3,...,50}. Exactly one of the ways to select these 6 numbers will result in the 6 winning numbers.

The probability is the number of favorable outcomes divided by the number of possible outcomes:

P(win)=(#of faverable outcome)/(#of possible outcomes)=1/C(50,6) =6!(50-6)!/50! =    =6!44!/50!=1/15,890,700

P(loss)=1−P(win)=1−1/15,890,700=15,890,699/15,890,700

When we win, then we gain $10 million and lose the cost of the lottery ticket ($10,000,000−1=$9,999,999).

When we lose, then we lose the cost of the lottery ticket (−$1).

The expected value (or mean) μ is the sum of the product of each possibility  with its probability P(x):

E(x)=∑xP(x)

=$9,999,999×1/15,890,700+(-$1)×15,890,699/15,890,700=-$15,890,699/15,890,700

=-$5,890,699/15,890,700=-58,907/158,907

≈−$0.37

Thus we are expected to lose $0.37 (or 37 cents) per lottery ticket.

Result

−$0.37