| --- |
| 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)). |
