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source: ncert
topic: probability
class_level: class_10
chapter: ch15
difficulty: intermediate
---
Class 10 | Ch15: Probability
CLASSICAL DEFINITION:
P(E) = Number of favourable outcomes / Total number of equally likely outcomes.
Range: 0 <= P(E) <= 1. P(E)+P(not E)=1.
P(certain event)=1. P(impossible event)=0.
SAMPLE SPACES:
One coin: S={H,T}, n(S)=2.
Two coins: S={HH,HT,TH,TT}, n(S)=4.
One die: S={1,2,3,4,5,6}, n(S)=6.
Two dice: n(S)=36 (all ordered pairs).
Deck of cards: n(S)=52. (4 suits × 13 cards each. Face cards=12. Aces=4.)
CARD PROBLEMS:
Red cards=26 (hearts+diamonds). Black=26 (clubs+spades).
Face cards=12 (J,Q,K of each suit). Aces=4.
King of red=2. Face card AND red=6.
TWO DICE PROBLEMS:
Sum=2: (1,1)=1 way. Sum=3: 2 ways. Sum=7: 6 ways (most common).
Sum=12: (6,6)=1 way.
P(sum=7)=6/36=1/6. P(sum=2 or 12)=2/36=1/18.
SOLVED EXAMPLES:
Example 1: Two dice. P(sum=8).
Pairs: (2,6),(3,5),(4,4),(5,3),(6,2)=5. P=5/36.
Example 2: Cards. P(king or red).
P(king)=4/52. P(red)=26/52. P(king AND red)=2/52.
P(king or red)=4/52+26/52-2/52=28/52=7/13.
Example 3: Bag: 5 red, 3 blue, 2 green. P(not red)?
P(not red)=(3+2)/10=5/10=1/2.
Example 4: Two dice. P(same number on both).
Pairs: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)=6. P=6/36=1/6.
COMMON MISTAKES:
Two coins: 3 outcomes (0H,1H,2H) are NOT equally likely. Use 4 outcomes (HH,HT,TH,TT).
P(A or B) = P(A)+P(B)-P(A and B). Don't forget the intersection.
For two dice: total outcomes=36 not 12 (order matters: (1,2)≠(2,1)). |