Theory of Probability Ex 16.3: Mastering Probability Concepts for Class 11

Understanding the Fundamentals of Probability

Probability is the measure of the likelihood that an event will occur. In simple terms, it tells us how likely something is to happen. It's expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. For example, the probability of a fair coin landing on heads is 0.5, meaning it's equally likely to happen or not happen.

Before we dive into calculations, let's clarify some key terms:

1. Random Experiment: An experiment whose outcome cannot be predicted with certainty, but all possible outcomes are known. For example, tossing a coin, rolling a die.
2. Sample Space (S): The set of all possible outcomes of a random experiment. For a coin toss, S = {Head, Tail}. For rolling a single die, S = {1, 2, 3, 4, 5, 6}.
3. Event (E): A subset of the sample space. It's a collection of one or more outcomes. For instance, getting an even number when rolling a die is an event E = {2, 4, 6}.
4. Elementary Event: An event having only one outcome of the random experiment. For example, getting a '3' when rolling a die is an elementary event.

Classical Definition of Probability:
The probability of an event E, denoted as P(E), is defined as:

P(E) = (Number of outcomes favourable to E) / (Total number of possible outcomes)

This formula is fundamental to solving problems in Exercise 16.3. Remember that for this definition to be applicable, all outcomes in the sample space must be equally likely. This means each outcome has an equal chance of occurring. This basic understanding forms the bedrock for solving the problems in this exercise and beyond.

Key Types of Events and Probability Rules

Impossible Event

An event that cannot occur. Its probability is 0. For example, getting an 8 when rolling a standard six-sided die. P(Impossible Event) = 0.

Sure (Certain) Event

An event that is certain to occur. Its probability is 1. For example, getting a number less than 7 when rolling a standard die. P(Sure Event) = 1.

Mutually Exclusive Events

Two events A and B are mutually exclusive if they cannot occur at the same time, meaning they have no common outcomes. If A and B are mutually exclusive, P(A or B) = P(A) + P(B). This is also written as P(A ∪ B) = P(A) + P(B).

Exhaustive Events

A set of events is exhaustive if at least one of them must occur whenever the experiment is performed. The union of all exhaustive events forms the entire sample space. For mutually exclusive and exhaustive events, the sum of their probabilities is 1.

Complementary Event

The complement of an event A, denoted as A' or A$^c$, is the event that A does not occur. The sum of the probability of an event and its complement is always 1: P(A) + P(A') = 1. Therefore, P(A') = 1 - P(A).

Solved Examples for Probability Calculations

• Example 1: Dice Roll Probability Question: A single fair die is rolled. What is the probability of: a) Getting an even number? b) Getting a number less than 3? Solution: Step 1: Identify the Sample Space (S). When a fair die is rolled, the possible outcomes are {1, 2, 3, 4, 5, 6}. So, n(S) = 6. Step 2: For part (a), identify the event E1 = 'getting an even number'. The outcomes favourable to E1 are {2, 4, 6}. So, n(E1) = 3. Step 3: Calculate P(E1). P(E1) = n(E1) / n(S) = 3 / 6 = 1/2. Step 4: For part (b), identify the event E2 = 'getting a number less than 3'. The outcomes favourable to E2 are {1, 2}. So, n(E2) = 2. Step 5: Calculate P(E2). P(E2) = n(E2) / n(S) = 2 / 6 = 1/3. Final Answer: The probability of getting an even number is 1/2. The probability of getting a number less than 3 is 1/3.
• Example 2: Card Drawing Probability Question: From a well-shuffled deck of 52 playing cards, one card is drawn at random. What is the probability that the card drawn is: a) A King? b) A red card? c) A face card? Solution: Step 1: Identify the Sample Space (S). The total number of cards in a well-shuffled deck is 52. So, n(S) = 52. Step 2: For part (a), identify the event E1 = 'drawing a King'. There are 4 Kings in a deck (King of Spades, King of Hearts, King of Diamonds, King of Clubs). So, n(E1) = 4. Step 3: Calculate P(E1). P(E1) = n(E1) / n(S) = 4 / 52 = 1/13. Step 4: For part (b), identify the event E2 = 'drawing a red card'. There are 26 red cards in a deck (13 Hearts + 13 Diamonds). So, n(E2) = 26. Step 5: Calculate P(E2). P(E2) = n(E2) / n(S) = 26 / 52 = 1/2. Step 6: For part (c), identify the event E3 = 'drawing a face card'. Face cards include King, Queen, and Jack. Each suit has 3 face cards, and there are 4 suits. So, n(E3) = 3 * 4 = 12. Step 7: Calculate P(E3). P(E3) = n(E3) / n(S) = 12 / 52 = 3/13. Final Answer: The probability of drawing a King is 1/13. The probability of drawing a red card is 1/2. The probability of drawing a face card is 3/13.

Exam Tips and Avoiding Common Mistakes in Probability

To excel in probability problems, especially those from Ex 16.3, keep these points in mind:

1. Clearly Define Sample Space: Always start by listing or determining the total number of possible outcomes (n(S)). This is the denominator in your probability calculation. A common mistake is to miscount this. For two dice, n(S) = 36, not 12. For two coins, n(S) = 4, not 2.
2. Identify Favourable Outcomes Precisely: Carefully read the event description and list all outcomes that satisfy the condition. Ensure no outcome is missed or double-counted. For example, in card problems, distinguish between 'red cards' and 'hearts'.
3. Simplify Fractions: Always present your final probability as a fraction in its simplest form (e.g., 2/4 should be 1/2).
4. Probability Range Check: Your calculated probability must always be between 0 and 1 (inclusive). If you get a value outside this range, recheck your calculations immediately.
5. Understanding 'OR' and 'AND': For mutually exclusive events, P(A or B) = P(A) + P(B). If events are not mutually exclusive (they can happen together), you need to subtract the probability of their intersection: P(A or B) = P(A) + P(B) - P(A and B). While Ex 16.3 primarily focuses on simpler cases, understanding this distinction is vital for future exercises.
6. Complementary Events: Don't forget that P(E') = 1 - P(E). This can simplify calculations, especially when it's easier to find the probability of an event not happening than of it happening directly.

Practice Questions with Solutions

• Q: A bag contains 3 red balls, 5 black balls, and 2 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: a) red? b) not white? A: Step 1: Determine the total number of balls (sample space). Total balls = 3 (red) + 5 (black) + 2 (white) = 10 balls. So, n(S) = 10. Step 2: For part (a), find the number of red balls (favourable outcomes). Number of red balls = 3. So, n(E_red) = 3. Step 3: Calculate the probability of drawing a red ball. P(red) = n(E_red) / n(S) = 3 / 10. Step 4: For part (b), find the number of white balls. Number of white balls = 2. So, n(E_white) = 2. Step 5: Calculate the probability of drawing a white ball. P(white) = n(E_white) / n(S) = 2 / 10 = 1/5. Step 6: Use the complementary event concept for 'not white'. P(not white) = 1 - P(white) = 1 - 1/5 = 4/5. Final answer: The probability of drawing a red ball is 3/10. The probability of drawing a ball that is not white is 4/5.
• Q: Two fair coins are tossed simultaneously. What is the probability of getting: a) Exactly one head? b) At least one tail? A: Step 1: Determine the sample space (S). When two coins are tossed, the possible outcomes are {HH, HT, TH, TT}. So, n(S) = 4. Step 2: For part (a), identify the event E1 = 'exactly one head'. The outcomes with exactly one head are {HT, TH}. So, n(E1) = 2. Step 3: Calculate P(E1). P(exactly one head) = n(E1) / n(S) = 2 / 4 = 1/2. Step 4: For part (b), identify the event E2 = 'at least one tail'. 'At least one tail' means one tail or two tails. The outcomes are {HT, TH, TT}. So, n(E2) = 3. Step 5: Calculate P(E2). P(at least one tail) = n(E2) / n(S) = 3 / 4. Final answer: The probability of getting exactly one head is 1/2. The probability of getting at least one tail is 3/4.
• Q: A letter is chosen at random from the letters of the word 'MATHEMATICS'. What is the probability that the letter chosen is a vowel? A: Step 1: Determine the total number of letters in the word 'MATHEMATICS'. The letters are M, A, T, H, E, M, A, T, I, C, S. There are 11 letters. So, n(S) = 11. Step 2: Identify the vowels in the word. The vowels are A, E, A, I. (Counting repetitions). So, the number of favourable outcomes (vowels) is 4. n(E_vowel) = 4. Step 3: Calculate the probability of choosing a vowel. P(vowel) = n(E_vowel) / n(S) = 4 / 11. Final answer: The probability that the letter chosen is a vowel is 4/11.
• Q: In a class of 30 students, 12 study English, 10 study Hindi, and 5 study both. If a student is chosen at random, what is the probability that the student studies: a) English or Hindi? b) Neither English nor Hindi? A: Step 1: Define events and given probabilities. Let E be the event that a student studies English, and H be the event that a student studies Hindi. Total students n(S) = 30. n(E) = 12, n(H) = 10, n(E ∩ H) = 5 (students studying both). Step 2: For part (a), calculate P(E or H) using the addition rule. n(E U H) = n(E) + n(H) - n(E ∩ H) = 12 + 10 - 5 = 17. P(E or H) = n(E U H) / n(S) = 17 / 30. Step 3: For part (b), calculate the number of students studying neither English nor Hindi. Number of students studying neither = Total students - n(E U H) = 30 - 17 = 13. Step 4: Calculate the probability of studying neither English nor Hindi. P(neither English nor Hindi) = 13 / 30. Final answer: The probability that the student studies English or Hindi is 17/30. The probability that the student studies neither is 13/30.

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