### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

# 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       =  $\frac{Number\: of\: favourable\: outcomes}{Possible\: outcomes}$
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)

## PROBABILITY WITH PLAYING CARDS

 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

## PROBABILITY WITH  TOSSING OF COINS

When one coin is tossed then sample space
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}

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}
 Formula for finding the sample space when number of coins are tossed = $2^{n}$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),
(3,1), 3,2), (3,3), 3,4), 3,5), 3,6),
(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
Number of doublets in this sample space = 6

{(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 = $6^{n}$Where n is the number of trials

Where  n is the number of trials

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.

## 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

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)

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