Probability — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Probability" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Probability" — important questions with detailed answers, download PDF for boa…
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Q1: A fair die is thrown once. What is the probability of getting a number greater than 4?
Step 1: Total possible outcomes when a die is thrown = {1, 2, 3, 4, 5, 6} = 6
Step 2: Favorable outcomes (number > 4) = {5, 6} = 2
Step 3: Probability = Favorable outcomes / Total outcomes
Step 4: P(number > 4) = 2/6 = 1/3
Final Answer: Probability = 1/3 or approximately 0.333
Q2: Two fair dice are thrown simultaneously. Find the probability of getting a sum of 7.
Step 1: Total possible outcomes when two dice are thrown = 6 × 6 = 36
Step 2: Favorable outcomes (sum = 7): (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Step 3: Number of favorable outcomes = 6
Step 4: Probability = 6/36 = 1/6
Final Answer: Probability = 1/6 or approximately 0.167
Q3: A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random. What is the probability of drawing a blue ball?
Step 1: Total number of balls = 5 + 3 + 2 = 10
Step 2: Number of blue balls = 3
Step 3: Probability = Number of blue balls / Total number of balls
Step 4: P(blue) = 3/10
Final Answer: Probability = 3/10 or 0.3
Q4: A card is drawn from a standard deck of 52 playing cards. What is the probability of drawing a face card (Jack, Queen, or King)?
Step 1: Total number of cards = 52
Step 2: Number of face cards in a deck:
Jacks: 4 (one of each suit)
Queens: 4 (one of each suit)
Kings: 4 (one of each suit)
Total face cards = 4 + 4 + 4 = 12
Step 3: Probability = Favorable outcomes / Total outcomes
Step 4: P(face card) = 12/52 = 3/13
Final Answer: Probability = 3/13 or approximately 0.231
Q5: A bag contains 4 red balls and 6 white balls. Two balls are drawn one after the other without replacement. Find the probability that both are red.
Step 1: Total number of balls = 4 + 6 = 10
Step 2: Probability of first ball being red = 4/10
Step 3: After drawing one red ball, remaining balls = 9
Step 4: Remaining red balls = 3
Step 5: Probability of second ball being red = 3/9
Step 6: Since both events are dependent:
P(both red) = P(first red) × P(second red | first red)
P(both red) = (4/10) × (3/9)
Step 7: P(both red) = 12/90 = 2/15
Final Answer: Probability = 2/15 or approximately 0.133
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