Probability: Cards, Dice, and Geometric Probability Master Guide Class 10

From assessing quality-control defect rates in industrial manufacturing to calculating the odds of a winning poker hand or modeling meteorology, probability theory is the mathematics of quantified chance. In CBSE Class 10 Mathematics, Chapter 14 (Probability) carries between $4$ and $6$ marks.

While elementary questions involve simple coin flips, high-scoring board exam problems test complex multi-factor scenarios: rolling two dice simultaneously (36 outcomes), navigating the taxonomy of a 52-card deck, solving conditional non-replacement problems, and computing continuous 2D geometric probabilities.

In this master guide, we synthesize the four major pillars of Class 10 probability into an authoritative reference.

What You Will Learn

• Theoretical Probability definition: $P(E) = \frac{n(E)}{n(S)}$
• The Two-Dice $36$-Outcome Grid: Sums ($2$ to $12$), doublets, and "5 will not come up either time"
• The 52-Card Deck Taxonomy: Suits, colours, the 12 face cards, and card-removal contractions
• Non-Replacement Quality Control: How sample space shrinks ($20 \to 19$)
• Continuous Geometric Probability: 2D area models (helicopter lake crash and dartboards)
• Common linguistic traps: "At least" vs. "At most"

1. The Two-Dice $36$-Outcome Matrix

When two dice are thrown together: $\mathbf{n(S) = 6 \times 6 = 36}$

           (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)

Key High-Frequency Probability Queries:

1. Doublets (Both dice show same number): $\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\} \implies \mathbf{P = \frac{6}{36} = \frac{1}{6}}$
2. Sum of Two Numbers is 7 (Most Probable Sum): $\{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \implies \mathbf{P = \frac{6}{36} = \frac{1}{6}}$
3. "5 Will Come Up At Least Once": 6 outcomes with 5 on first die $+ 6$ outcomes with 5 on second die $- 1$ overlap $(5, 5)$: $n(E) = 6 + 6 - 1 = 11 \implies \mathbf{P = \frac{11}{36}}$
4. "5 Will Not Come Up Either Time": Complementary event: $1 - \frac{11}{36} = \mathbf{\frac{25}{36}}$.

2. The 52-Card Deck Taxonomy

                                  52 Playing Cards
                                         |
       +---------------------------------+---------------------------------+
       |                                                                   |
26 RED CARDS                                                        26 BLACK CARDS
- 13 Hearts (♥)                                                     - 13 Spades (♠)
- 13 Diamonds (♦)                                                   - 13 Clubs (♣)
       |                                                                   |
6 RED FACE CARDS                                                    6 BLACK FACE CARDS
(2 Kings, 2 Queens, 2 Jacks)                                        (2 Kings, 2 Queens, 2 Jacks)

The Face Card Truth (CBSE Core Focus): <u>There are exactly 12 FACE CARDS in a deck (Kings, Queens, Jacks). ACES ARE NOT FACE CARDS! Aces are honour cards. Counting aces as face cards is the most common student error in board exams!</u>

Solved Example: Card Removal Problem

Problem: From a deck of $52$ cards, all four kings are removed. One card is then drawn at random. Find the probability that the card drawn is: (i) a face card, (ii) a black card.

Solution:

1. Four kings are removed $\implies$ New total sample space: $n(S') = 52 - 4 = \mathbf{48\text{ cards}}$
2. (i) A face card: Originally $12$ face cards. After removing $4$ kings, $8$ face cards remain (4 Queens, 4 Jacks): $P(\text{Face Card}) = \frac{8}{48} = \mathbf{\frac{1}{6}}$
3. (ii) A black card: Originally $26$ black cards. Two black kings were removed $\implies 26 - 2 = 24$ black cards remain: $P(\text{Black Card}) = \frac{24}{48} = \mathbf{\frac{1}{2}}$
4. Therefore, <u>the probability of a face card is rac{1}{6} and of a black card is rac{1}{2}</u>.

3. Geometric Probability (Continuous 2D Area Models)

When outcomes are infinite points scattered uniformly across a two-dimensional surface: $\mathbf{P(E) = \frac{\text{Area of Favourable Target Region}}{\text{Total Area of Sample Space Region}}}$

Solved Example: The Circular Target in a Rectangle

Problem: A dart is dropped at random onto a rectangular region of dimensions $3\text{ m} \times 2\text{ m}$. Inside the rectangle, a circular target of diameter $1\text{ m}$ is drawn. What is the probability that the dart lands inside the circle?

    +--------------------------- 3 m ---------------------------+
    |                                                           |
    |                     ( Circle d = 1 m )                    | 2 m
    |                                                           |
    +-----------------------------------------------------------+

Solution:

1. Total Area of Rectangular Region: $A_{\text{total}} = \text{Length} \times \text{Breadth} = 3\text{ m} \times 2\text{ m} = \mathbf{6\text{ m}^2}$
2. Area of Circular Target: Diameter $d = 1\text{ m} \implies$ Radius $r = 0.5\text{ m} = \frac{1}{2}\text{ m}$. $A_{\text{circle}} = \pi r^2 = \pi \times \left(\frac{1}{2}\right)^2 = \mathbf{\frac{\pi}{4}\text{ m}^2}$
3. Apply Geometric Probability: $P(\text{Inside Circle}) = \frac{A_{\text{circle}}}{A_{\text{total}}} = \frac{\frac{\pi}{4}}{6} = \mathbf{\frac{\pi}{24}}$
4. Therefore, <u>the probability that the dart lands inside the circle is rac{\pi}{24}</u>.

4. Summary and Examination Tips

Exam Tip: In questions asking for the probability that two friends have the same birthday in a non-leap year: Favourable days $= 1 \implies P = \mathbf{\frac{1}{365}}$. For different birthdays: $P = 1 - \frac{1}{365} = \mathbf{\frac{364}{365}}$!

Common Mistake: In non-replacement problems, calculating with the original sample space. If a bulb or card is not replaced, reduce the denominator by $1$!