# Probability Math Basic Concept

Random Experiment:

An experiment in which all possible outcomes are known and the exact output cannot be predicted in advance is called a random

experiment.

Sample Space

When we perform an experiment, then the set S of all possible outcomes is called the sample space.

If we through a dice, the sample space

S={1, 2, 3, 4, 5, 6}

Event

Any subset of a sample space is called an event. After tossing a coin, getting Head on the top is an event.

Probability of an event

•Let S be the sample and let E be an event.

•Then E ⊆S

•Probability of E for occurring, P(E)= ( n(E)) / (n(S) )

•Here n(E) is the number of elements in E and n(s) is the number of elements in S.

Probability Axioms

•The probability of an event always varies from 0 to 1. [0≤P(x)≤1]

•For an impossible event the probability is 0 and for a certain event the

probability is 1.

•If the occurrence of one event is not influenced by another event, they are called

mutually exclusive or disjoint. If A1,A2….An are mutually exclusive/disjoint

events, then

P(A1∪A2∪….An)=P(A1)+P(A2)+…..P(An)

Properties of Probability

● If there are two events x and x̅ which are complementary, then the

probability of the complementary event is −

p(x̅ )=1−p(x)

A Standard Deck of Card

Examples

Example 1: What is the probability of drawing a king from a deck of cards?

Solution 1:

Here the event E is drawing a king from a deck of cards.

There are 52 cards in a deck of cards.

Hence, total number of outcomes = 52

The number of favorable outcomes = 4 (as there are 4 kings in a deck)

Hence, the probability of this event occurring is

P(E) = 4/52 = 1/13

Examples

Example 2: Andy has drawn a card from a well-shuffled deck. Help her find the

probability of the card either being red or a King.

Solution 2:

Jessica knows here that event E is the card drawn being either red or a King.

The total number of outcomes = 52

There are 26 red cards, and 4 cards which are Kings.

However, 2 of the red cards are Kings.

Examples

If we add 26 and 4, we will be counting these two cards twice.

Thus, the correct number of outcomes which are favorable to E is

26 + 4 – 2 = 28

Hence, the probability of event occurring is

P(E) = 28/52 = 7/13

Examples

Example 3: Two cards are drawn from the pack of 52 cards. Find the probability

that both are diamonds or both are kings.

Solution 3:

Total no. of ways = 52C2

Case I: Both are diamonds = 13C2

Case II: Both are kings = 4C2

P (both are diamonds or both are kings) = (13C2 + 4C2 ) / 52C2

Examples

Example 4: A problem is given to three persons P, Q, R whose respective chances

of solving it are 2/7, 4/7, 4/9 respectively. What is the probability that the

problem is solved?

Solution 4:

Probability of the problem getting solved = 1 – (Probability of none of them

solving the problem)

Probability of problem getting solved = 1 – (5/7) x (3/7) x (5/9) = (122/147)

Examples

Example 5: There are 5 green 7 red balls. Two balls are selected one by one

without replacement. Find the probability that first is green and second is red.

Solution 5:

P (G) × P (R) = (5/12) x (7/11) = 35/132

Examples

Example 6: Two cards are drawn from the pack of 52 cards. Find the probability

that both are diamonds or both are kings.

Solution 6:

Total no. of ways = 52C2

Case I: Both are diamonds = 13C2

Case II: Both are kings = 4C2

P (both are diamonds or both are kings) = (13C2 + 4C2 ) / 52C2

Examples

Example 7: Find the probability that a leap year has 52 Sundays.

Solution 7:

A leap year can have 52 Sundays or 53 Sundays. In a leap year, there are 366

days out of which there are 52 complete weeks & remaining 2 days. Now,

these two days can be (Sat, Sun) (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed,

Thur) (Thur, Friday) (Friday, Sat).

So there are total 7 cases out of which (Sat, Sun) (Sun, Mon) are two

favorable cases. So, P (53 Sundays) = 2 / 7

Examples

Now, P(52 Sundays) + P(53 Sundays) = 1

So, P (52 Sundays) = 1 – P(53 Sundays) = 1 – (2/7) = (5/7)

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