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)