CBSE Class 10 Maths: Probability

Understanding Probability: The Mathematics of Chance

Probability is a branch of mathematics that deals with the likelihood of an event occurring. In simpler terms, it's a measure of how likely something is to happen. For your probability Class 10 NCERT studies, we use a numerical value between 0 and 1 to represent this likelihood.

To understand probability, we first need to define a few key terms:

1. Experiment: An operation that produces some well-defined outcomes. For example, tossing a coin, rolling a die, or drawing a card from a deck are all experiments.
2. Outcome: A possible result of an experiment. When you toss a coin, 'Head' is an outcome, and 'Tail' is another outcome.
3. Sample Space (S): The set of all possible outcomes of an experiment. For tossing a coin, the sample space S = {Head, Tail}. For rolling a standard six-sided die, S = {1, 2, 3, 4, 5, 6}.
4. Event (E): A subset of the sample space; it is a collection of one or more outcomes from an experiment. For instance, in rolling a die, 'getting an even number' is an event, consisting of outcomes {2, 4, 6}.

The probability of an event E, denoted as P(E), is calculated using the formula:

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

This formula is central to solving almost all probability problems in Class 10. Remember, the outcomes must be equally likely for this formula to be accurate.

Key Types of Events in Probability

Elementary Event

An event having only one outcome of the experiment. For example, when rolling a die, getting a '2' is an elementary event, as it consists of only one outcome.

Compound Event

An event having more than one outcome. For example, when rolling a die, 'getting an even number' is a compound event, as it includes outcomes {2, 4, 6}.

Sure (Certain) Event

An event that is certain to happen. Its probability is always 1. For example, 'getting a number less than 7' when rolling a standard die is a sure event.

Impossible Event

An event that cannot happen. Its probability is always 0. For example, 'getting an 8' when rolling a standard die is an impossible event.

Complementary Events

If E is an event, then 'not E' (denoted as E' or Ē) is its complementary event. The sum of the probabilities of an event and its complementary event is always 1, i.e., P(E) + P(E') = 1. For example, if E is 'getting a Head' in a coin toss, E' is 'getting a Tail'.

Step-by-Step Guide to Calculating Probability

1. Step 1: Identify the Experiment and Sample Space — Clearly understand the experiment being performed. List all possible outcomes to form the Sample Space (S). This is crucial for determining the total number of possible outcomes, n(S).
2. Step 2: Determine the Total Number of Outcomes, n(S) — Count the total number of distinct outcomes in your Sample Space (S). Ensure all outcomes are equally likely. For example, if you roll a fair die, n(S) = 6.
3. Step 3: Identify the Event (E) and Favourable Outcomes — Read the question carefully to understand what specific event (E) you need to find the probability of. List all the outcomes from the sample space that satisfy the conditions of Event E. These are your favourable outcomes.
4. Step 4: Count the Number of Favourable Outcomes, n(E) — Count the number of outcomes that belong to Event E. This is n(E). For example, if Event E is 'getting an even number' when rolling a die, the favourable outcomes are {2, 4, 6}, so n(E) = 3.
5. Step 5: Apply the Probability Formula — Substitute the values of n(E) and n(S) into the probability formula: P(E) = n(E) / n(S). Simplify the resulting fraction to its lowest terms. For the die example, P(E) = 3/6 = 1/2.
6. Example: Probability of Getting a Prime Number on a Die Roll — Let's apply these steps. Experiment: Rolling a fair six-sided die. Step 1 & 2 (Sample Space & Total Outcomes): The sample space S = {1, 2, 3, 4, 5, 6}. So, n(S) = 6. Step 3 (Event & Favourable Outcomes): Let Event E be 'getting a prime number'. The prime numbers in S are {2, 3, 5}. Step 4 (Number of Favourable Outcomes): n(E) = 3. Step 5 (Apply Formula): P(E) = n(E) / n(S) = 3 / 6 = 1/2. Thus, the probability of getting a prime number when rolling a die is 1/2.

Worked Examples: Applying Probability Concepts

• Example 1: Drawing a Card A card is drawn at random from a well-shuffled deck of 52 playing cards. What is the probability of getting: (a) an Ace (b) a red card (c) a face card Solution: Total number of possible outcomes, n(S) = 52 (since there are 52 cards in a deck). (a) Event E: Getting an Ace Number of Aces in a deck = 4 (Ace of Spades, Ace of Clubs, Ace of Hearts, Ace of Diamonds). Number of favourable outcomes, n(E) = 4. P(E) = n(E) / n(S) = 4 / 52 = 1 / 13. (b) Event F: Getting a red card Number of red cards in a deck = 26 (13 Hearts + 13 Diamonds). Number of favourable outcomes, n(F) = 26. P(F) = n(F) / n(S) = 26 / 52 = 1 / 2. (c) Event G: Getting a face card Face cards include Jacks, Queens, and Kings. There are 4 of each suit. Number of face cards = 3 types * 4 suits = 12. Number of favourable outcomes, n(G) = 12. P(G) = n(G) / n(S) = 12 / 52 = 3 / 13.
• Example 2: Balls in a Bag A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (a) red (b) not red Solution: Total number of balls in the bag = 3 (red) + 5 (black) = 8. Total number of possible outcomes, n(S) = 8. (a) Event E: The ball drawn is red Number of red balls = 3. Number of favourable outcomes, n(E) = 3. P(E) = n(E) / n(S) = 3 / 8. (b) Event E': The ball drawn is not red 'Not red' means the ball is black. Number of black balls = 5. Number of favourable outcomes, n(E') = 5. P(E') = n(E') / n(S) = 5 / 8. Alternatively, using complementary events: P(E') = 1 - P(E) = 1 - (3/8) = 5/8.
• Example 3: Tossing Two Coins Two coins are tossed simultaneously. What is the probability of getting: (a) exactly one head (b) at least one head (c) no heads Solution: When two coins are tossed, the sample space S = {HH, HT, TH, TT}. Total number of possible outcomes, n(S) = 4. (a) Event E: Getting exactly one head Favourable outcomes = {HT, TH}. Number of favourable outcomes, n(E) = 2. P(E) = n(E) / n(S) = 2 / 4 = 1 / 2. (b) Event F: Getting at least one head 'At least one head' means one head or two heads. Favourable outcomes = {HH, HT, TH}. Number of favourable outcomes, n(F) = 3. P(F) = n(F) / n(S) = 3 / 4. (c) Event G: Getting no heads 'No heads' means both are tails. Favourable outcomes = {TT}. Number of favourable outcomes, n(G) = 1. P(G) = n(G) / n(S) = 1 / 4.

Exam Tips and Avoiding Common Mistakes in Probability

To excel in CBSE Class 10 Maths probability, pay attention to these critical points and common pitfalls:

• Probability Range: Always remember that the probability of any event E, P(E), must lie between 0 and 1, inclusive (i.e., 0 ≤ P(E) ≤ 1). If your calculated probability is negative or greater than 1, you've made a mistake.
• Sum of Probabilities: The sum of the probabilities of all elementary events of an experiment is always 1. This is a good way to cross-check your calculations.
• Equally Likely Outcomes: The basic probability formula P(E) = n(E)/n(S) is valid only when all outcomes in the sample space are equally likely. Ensure this condition is met.
• Careful Counting: Especially in problems involving cards, dice (single or multiple), or selecting items from a group, carefully list and count the total possible outcomes and favourable outcomes. A common mistake is miscounting.
• Complementary Events: Utilize the concept of complementary events (P(E') = 1 - P(E)) whenever it simplifies the calculation. For example, finding the probability of 'at least one head' can sometimes be easier by finding the probability of 'no heads' and subtracting from 1.
• Read the Question Thoroughly: Distinguish between 'exactly,' 'at least,' 'at most,' and 'not' in the question. Each word changes the set of favourable outcomes.

Practice Questions with Solutions

• Q: A die is thrown once. Find the probability of getting a number: (i) greater than 4 (ii) less than or equal to 4 A: Step 1: Identify the sample space. When a die is thrown, the possible outcomes are S = {1, 2, 3, 4, 5, 6}. Total number of outcomes, n(S) = 6. Step 2: For (i) Event E: getting a number greater than 4. Favourable outcomes = {5, 6}. Number of favourable outcomes, n(E) = 2. Step 3: Calculate P(E) = n(E) / n(S) = 2 / 6 = 1/3. Step 4: For (ii) Event F: getting a number less than or equal to 4. Favourable outcomes = {1, 2, 3, 4}. Number of favourable outcomes, n(F) = 4. Step 5: Calculate P(F) = n(F) / n(S) = 4 / 6 = 2/3. Final answer: (i) 1/3, (ii) 2/3
• Q: A box contains 5 red marbles, 8 white marbles, and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be: (i) red? (ii) white? (iii) not green? A: Step 1: Identify the total number of marbles. Total marbles = 5 + 8 + 4 = 17. Total number of outcomes, n(S) = 17. Step 2: For (i) Event E: marble is red. Number of red marbles = 5. n(E) = 5. Step 3: Calculate P(E) = n(E) / n(S) = 5 / 17. Step 4: For (ii) Event F: marble is white. Number of white marbles = 8. n(F) = 8. Step 5: Calculate P(F) = n(F) / n(S) = 8 / 17. Step 6: For (iii) Event G: marble is not green. This means it can be red or white. Number of non-green marbles = 5 (red) + 8 (white) = 13. n(G) = 13. Step 7: Calculate P(G) = n(G) / n(S) = 13 / 17. Final answer: (i) 5/17, (ii) 8/17, (iii) 13/17
• Q: A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting a King of red colour. A: Step 1: Identify the total number of cards. Total number of outcomes, n(S) = 52. Step 2: Identify the event E: getting a King of red colour. Red Kings are King of Hearts and King of Diamonds. There are 2 such cards. Step 3: Number of favourable outcomes, n(E) = 2. Step 4: Calculate P(E) = n(E) / n(S) = 2 / 52 = 1 / 26. Final answer: 1/26
• Q: What is the probability that a number selected from the numbers 1, 2, 3, ..., 25 is a prime number? A: Step 1: Identify the sample space. S = {1, 2, 3, ..., 25}. Total number of outcomes, n(S) = 25. Step 2: Identify the event E: selecting a prime number. The prime numbers between 1 and 25 are {2, 3, 5, 7, 11, 13, 17, 19, 23}. Step 3: Count the number of favourable outcomes, n(E) = 9. Step 4: Calculate P(E) = n(E) / n(S) = 9 / 25. Final answer: 9/25
• Q: A game consists of tossing a 1 rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result (i.e., three heads or three tails), and loses otherwise. Calculate the probability that Hanif will lose the game. A: Step 1: Identify the sample space for tossing a coin 3 times. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total number of outcomes, n(S) = 8. Step 2: Identify the event A: Hanif wins. This means getting three heads or three tails. Favourable outcomes for A = {HHH, TTT}. Number of favourable outcomes, n(A) = 2. Step 3: Calculate the probability that Hanif wins, P(A) = n(A) / n(S) = 2 / 8 = 1/4. Step 4: Identify the event A': Hanif loses. This is the complementary event to Hanif winning. P(A') = 1 - P(A). Step 5: Calculate P(A') = 1 - 1/4 = 3/4. Final answer: 3/4

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