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.
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Probability — Textbook Questions Answered (9)
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
Q6: Three coins are tossed simultaneously. What is the probability of getting at least one tail?
Step 1: Total possible outcomes = 2³ = 8
Step 2: All possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Step 3: Method 1 (Direct counting):
Favorable outcomes (at least one tail): HHT, HTH, HTT, THH, THT, TTH, TTT = 7
P(at least one tail) = 7/8
Step 4: Method 2 (Complement):
P(at least one tail) = 1 - P(no tail) = 1 - P(all heads)
P(all heads) = 1/8
P(at least one tail) = 1 - 1/8 = 7/8
Final Answer: Probability = 7/8 or 0.875
Q7: In a lottery, there are 10 tickets out of which 3 are winning tickets. If you buy 2 tickets, what is the probability that both are winning tickets?
Step 1: Total number of tickets = 10
Step 2: Winning tickets = 3
Step 3: We need to find P(both tickets are winning)
Step 4: Number of ways to choose 2 winning tickets from 3 = C(3,2) = 3
Step 5: Total number of ways to choose 2 tickets from 10 = C(10,2) = 10!/(2!×8!) = 45
Step 6: Probability = Favorable outcomes / Total outcomes
Step 7: P(both winning) = 3/45 = 1/15
Step 8: Alternatively using conditional probability:
P(first winning) = 3/10
P(second winning | first winning) = 2/9
P(both winning) = (3/10) × (2/9) = 6/90 = 1/15
Final Answer: Probability = 1/15 or approximately 0.067
Q8: A number is chosen at random from the numbers 1 to 30. What is the probability that the chosen number is a multiple of 5?
Step 1: Total numbers = 30
Step 2: Multiples of 5 from 1 to 30: 5, 10, 15, 20, 25, 30
Step 3: Count of multiples of 5 = 6
Step 4: Probability = Favorable outcomes / Total outcomes
Step 5: P(multiple of 5) = 6/30 = 1/5
Final Answer: Probability = 1/5 or 0.2
Q9: A box contains 6 balls numbered 1 to 6. Two balls are drawn one after the other with replacement. Find the probability that the sum of the numbers on the balls is 8.
Step 1: Total possible outcomes = 6 × 6 = 36 (with replacement)
Step 2: Favorable outcomes (sum = 8):
(2,6): sum = 8
(3,5): sum = 8
(4,4): sum = 8
(5,3): sum = 8
(6,2): sum = 8
Total favorable outcomes = 5
Step 3: Probability = 5/36
Final Answer: Probability = 5/36 or approximately 0.139
Related
- NCERT Solutions: Probability — the CBSE version of this same topic.
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