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Resource Centre Mathematics Mathematics worksheet, mathematics basic points and formulas, mathematics lesson plan, mathematics multiple choice questions Workplace Dashboard CBSE Syllabus For Session 202324 For : Classes IX & X  Classes XI & XII Watch Videos on Maths Solutions CLASS IX MATHEMATICS FORMULAS & BASIC CONCEPTS
PROBABILITY Part1
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PROBABILITY
For classes 8th, 9th, 10th, 11th
Basic concepts of probability useful for classes 9, 10 and 11. Explanation with playing cards, tossing one coin, two coins and three coins.
Probability of an event = 
If E is an event then : P(E) + P(notE) = 1
Probability of an event is greater than, equal to 0 but is less than, equal to 1. This is also be called as the range of Probability
\[0\leq P(E)\leq 1\] 
Impossible Event:
An event which never occur is called impossible event.
P(Impossible Event) = 0
Sure Event :
An even which happens for sure is called sure event. Sure event is also be called as the certain event.
P(Sure Event or Certain Event) = 1
Equally likely events:
If every event of the sample space has equal possibility to happen, then the events are called equally likely events. (e.g. Tossing a coin)
DESCRIPTION OF PLAYING
CARDS 

Total playing cards 
52 
Number of black cards 
26 
Number of red cards 
26 
Number of spade card (♠) 
13 
Number of Clubs card(♣) 
13 
Number of diamond card(⯁) 
13 
Number of heart card(❤) 
13 
Number of king card 
4 
Number of queen card 
4 
Number of Jack card 
4 
Number of ace card 
4 
Number of face card (Jack, Queen,
King) 
12 
S = {H, T} ⇒ n(S) = 2
When two coins are tossed then sample space
S = {HH, HT, TH, TT} ⇒ n(S) = 4 = {1, 2, 1}
When three coins are tossed then sample space
S = {HHH,HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ n(S) = 8 = {1, 3, 3, 1}
⇒ n(S) = 8 = {1, 3, 3, 1}
When three coins are tossed then sample space
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, THTH, TTHH, HTTH, HTTT, THTT, TTHT, TTTH, TTTT}
⇒ n(S) = 16 = {1, 4, 6, 4, 1}
⇒ n(S) = 16 = {1, 4, 6, 4, 1}
Formula for finding the sample space when number of coins are tossed = Where n is the number of trials 
PROBABILITY WITH TOSSING OF DICE
When a die is thrown then sample space
S = {1,2,3,4,5,6} ⇒ n(S) = 6
When two dice are thrown then sample space
S = {(1,1), (1,2), 1,3), 1,4), 1,5), 1,6),
(2,1), (2,2), 2,3), 2,4), 2,5), 2,6),
(2,1), (2,2), 2,3), 2,4), 2,5), 2,6),
(3,1), 3,2), (3,3), 3,4), 3,5), 3,6),
(4,1), (4,2), 4,3), 4,4), 4,5), 4,6),
(4,1), (4,2), 4,3), 4,4), 4,5), 4,6),
(5,1), 5,2), 5,3), 5, 4), 5, 5), 5, 6),
(6,1), (6,2), 6,3), (6,4), 6,5), 6,6)}
⇒ n(S) = 36
(6,1), (6,2), 6,3), (6,4), 6,5), 6,6)}
⇒ n(S) = 36
Number of doublets in this sample space = 6
{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} are called doublets
{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} are called doublets
When three dice are thrown then.
Number of elements in sample space = n(S) = 216
Formula for finding the sample space when number of coins are tossed = Where n is the number of trials 
Where n is the number of trials
Al least ⇒ ≥ (greater than or equal to)
Al least ⇒ ≥ (greater than or equal to)
At most ⇒ ≤ (less than or equal to)
If there is "or" between two or more events then the probability of different events are added.
If there is "and" between two or more events, then first we find the common events and then required probability is the probability of common events.
Solution: This question can be explained from the following table
Important Explanation
Question: When two dice are thrown then what is the probability such that sum of numbers on two dice is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.Solution: This question can be explained from the following table
Sum of the
numbers on two dice

Observations

No. of
observations

Probability

2

(1,1)

1

1/36

3

(1,2), (2,1)

2

2/36

4

(1,3), (2,2), (3,1)

3

3/36

5

(1,4), (2, 3), (3,2), (4,1)

4

4/36

6

(1,5), (2,4), (3,3), (4,2),
(5,1)

5

5/36

7

(1,6), (2,5), (3,4), (4,3),
(5,2), (6,1)

6

6/36

8

(2.6), (3,5), (4,4), (5,3),
(6,2)

5

5/36

9

(3, 6),(4, 5), (5,4), (6, 3)

4

4/36

10

(4,6), (5,5), (6,4)

3

3/36

11

(5,6), (6,5)

2

2/36

12

(6,6)

1

1/36

Two solve this problem first we write all 36 observations obtained after throwing two dice, then find the number of observations having sum 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, then find their probability.
This can be made easy if we follow the figure below. we understand it as
This can be made easy if we follow the figure below. we understand it as
No of observations with sum 2 = 1
P( Getting sum 2) = 1/36
No of observations with sum 3 = 2
P( Getting sum 3) = 2/36
No of observations with sum 7 = 6
P( Getting sum 7) = 6/36
No of observations with sum 11 = 2
P( Getting sum 11) = 2/36
........................ and so on
Next topics are for 10+1 and 10+2 class students
Simple event
If an event have only one sample point of sample space, it is called a simple event.
For example : HH, HHH, TT, TTT, 222, all are called simple events.
Compound Event:
If an event have more than one sample point in a sample space then it is called a compound event.
For Example: HT, THH, 234, etc all are compound event.
Mutually Exclusive events:
If A and B are two events and A⋂B = Φ, then A and B are called mutually exclusive events.
Exhaustive events:
If union of all the events is equal to the sample space then the events are called exhaustive events.
Consistency of Probability:
If A and B are two events then probability is said to be consistently defined if
P(AUB) > P(A) and P(AUB) > P(B)
Other Important Results on Probability
P(A or B) = P(A U B) = P(A) + P(B)  P(A⋂ B)
P(Either A or B) = P(AUB)
P(A and B) = P(A ⋂ B)
P(Neither A nor B) = P(A' ⋂ B') = P(AUB)' = 1  P(AUB)
P(not A) = P(A') = 1 P(A)
P(not A or not B) = P(A' U B') = P(A ⋂ B)' = 1  P(A ⋂ B)
P(A⋂B') = P(A  B) = P(A)  P(A⋂B)
P(A'⋂B) = P(B  A) = P(B)  P(A⋂B)
If A and B are two events then P(At least one of them ) = P(AUB)
INTEGRATION OF PROBABILITY WITH ARTIFICIAL INTELLIGENCE TOOL
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