Probability Ex 15.2 Class 10 NCERT Solutions

Understanding the Concepts Behind Ex 15.2

Exercise 15.1 laid a strong foundation by introducing the basic formula for theoretical probability: P(E) = (Number of favourable outcomes) / (Total number of outcomes). The problems were straightforward, dealing with single events like tossing a coin, rolling a die, or drawing a card.

Exercise 15.2 takes this a step further. The problems are not designed to trick you, but to make you think more critically about what constitutes 'total outcomes' and 'favourable outcomes'. Here, you'll encounter:

1. Compound Events: Instead of one die, what if you roll two? Instead of one friend's birthday, what about two? These situations require you to systematically list or calculate all possible combined outcomes. For two dice, the total outcomes aren't 6+6=12, but 6x6=36.
2. Complementary Events: Many questions are easier to solve by finding the probability of the event not happening. For example, finding the probability of 'at least one head' in three coin tosses is complex. It's much simpler to find the probability of 'no heads' (which is just TTT) and subtract it from 1. This uses the rule: P(E) = 1 - P(not E).
3. Geometric Probability: Some questions involve shapes. For instance, what's the probability of a dart, thrown randomly, landing inside a smaller circle within a larger one? Here, outcomes aren't counted; they are measured by area. The formula becomes P(E) = (Favourable Area) / (Total Area). This introduces a beautiful link between geometry and probability.

Worked Examples for Clarity

• Problem: Two friends, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability that Reshma will win the match? Solution: Step 1: Identify the events. Let S be the event that Sangeeta wins and R be the event that Reshma wins. In a tennis match, one person must win; there are no ties. Therefore, the event 'Reshma wins' is the complement of the event 'Sangeeta wins'. Step 2: Use the complementary event formula. The sum of probabilities of an event and its complement is 1. P(S) + P(R) = 1 Step 3: Substitute the known value and solve. We are given P(S) = 0.62. 0.62 + P(R) = 1 P(R) = 1 - 0.62 P(R) = 0.38 Final Answer: The probability that Reshma will win the match is 0.38.
• Problem: A die is thrown twice. What is the probability that 5 will not come up either time? Solution: Method 1: Using Complementary Events (Easier) Step 1: Find the probability of the opposite event. Let's find the probability that 5 will come up at least once. It's easier to first find the probability that 5 will not come up on a single throw. P(not 5 on one throw) = 5/6. Step 2: Calculate the probability for two throws. Since the throws are independent, the probability that 5 will not come up on either throw is the product of the individual probabilities. P(5 not in 1st throw AND 5 not in 2nd throw) = P(not 5 on one throw) × P(not 5 on one throw) = (5/6) × (5/6) = 25/36. Method 2: Counting Outcomes (More Tedious) Step 1: Find the total number of outcomes. When a die is thrown twice, the total number of possible outcomes is 6 × 6 = 36. Step 2: Find the number of favourable outcomes. We want outcomes where 5 does not appear. For the first throw, there are 5 possibilities (1, 2, 3, 4, 6). For the second throw, there are also 5 possibilities. The total number of favourable outcomes is 5 × 5 = 25. Step 3: Calculate the probability. P(5 will not come up) = (Number of favourable outcomes) / (Total number of outcomes) = 25 / 36. Final Answer: The probability that 5 will not come up either time is 25/36.

Common Pitfalls to Avoid

• Confusing 'And' with 'Or': When an event requires 'A and B' to happen, you often multiply probabilities (for independent events). When it requires 'A or B', you often add probabilities. Don't mix these up.
• Incorrectly Identifying Total Outcomes: For two dice, the total outcomes are 36, not 12. For three coins, it's 2³ = 8, not 3×2=6. Always ensure you have the correct sample space.
• Assuming All Outcomes are Equally Likely: A classic trick question: "A player throws a die. What is the probability of throwing a 6 or not a 6?" Some might say 1/2 because there are two outcomes. This is wrong! The outcome '6' has a probability of 1/6, and 'not a 6' has a probability of 5/6. These are not equally likely outcomes.
• Forgetting P(E) = 1 - P(not E): For problems asking for 'at least one', it's almost always easier to calculate the probability of 'none' and subtract it from 1. Trying to calculate the possibilities for 'at least one' directly can be long and prone to errors.

Practice Questions with Solutions

• Q: Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is a prime number? A: Step 1: Determine the total number of outcomes. When two dice are thrown, the total number of possible outcomes is 6 × 6 = 36. Step 2: Identify the favourable outcomes. The possible sums range from 2 to 12. The prime numbers in this range are 2, 3, 5, 7, 11. - Sum = 2: (1,1) -> 1 outcome - Sum = 3: (1,2), (2,1) -> 2 outcomes - Sum = 5: (1,4), (4,1), (2,3), (3,2) -> 4 outcomes - Sum = 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) -> 6 outcomes - Sum = 11: (5,6), (6,5) -> 2 outcomes Total number of favourable outcomes = 1 + 2 + 4 + 6 + 2 = 15. Step 3: Calculate the probability. P(sum is prime) = (Favourable Outcomes) / (Total Outcomes) = 15/36 = 5/12. Final answer: 5/12.
• Q: A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag. A: Step 1: Set up the variables. Let the number of blue balls be 'x'. Total number of balls = 5 + x. Probability of drawing a red ball, P(R) = 5 / (5 + x). Probability of drawing a blue ball, P(B) = x / (5 + x). Step 2: Use the condition given in the question. We are given that P(B) = 2 P(R). Step 3: Formulate and solve the equation. x / (5 + x) = 2 [5 / (5 + x)] x / (5 + x) = 10 / (5 + x) Since the denominators are the same (and not zero), we can equate the numerators: x = 10. Final answer: There are 10 blue balls in the bag.
• Q: What is the probability that a leap year, selected at random, will contain 53 Sundays? A: Step 1: Analyze the structure of a leap year. A leap year has 366 days. 366 days = 52 weeks and 2 extra days. (366 ÷ 7 = 52 remainder 2). Step 2: Identify the sample space for the two extra days. These two extra days can be: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). There are a total of 7 possible pairs of consecutive days. Step 3: Find the favourable outcomes. For the year to have 53 Sundays, one of these two extra days must be a Sunday. The favourable pairs are (Sunday, Monday) and (Saturday, Sunday). There are 2 favourable outcomes. Step 4: Calculate the probability. P(53 Sundays) = (Favourable Outcomes) / (Total Outcomes) = 2/7. Final answer: 2/7.
• Q: A circular dartboard has a radius of 14 cm. Inside it, there is a circular bullseye with a radius of 7 cm. If a dart is thrown and it hits the board, what is the probability that it hits the bullseye? A: Step 1: Understand the principle of geometric probability. In this case, Probability = (Favourable Area) / (Total Area). Step 2: Calculate the total area. This is the area of the entire dartboard. Area of a circle = πr². Total Area = π (14)² = 196π cm². Step 3: Calculate the favourable area. This is the area of the bullseye. Favourable Area = π (7)² = 49π cm². Step 4: Calculate the probability. P(hitting the bullseye) = (Favourable Area) / (Total Area) = (49π) / (196π). The π cancels out. P = 49 / 196. Simplifying the fraction (dividing numerator and denominator by 49), we get 1/4. Final answer: 1/4.

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