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Chapter 14: Probability

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This page provides comprehensive Chapter 14: Probability - HOTS Worksheet - SJMaths. High Order Thinking Skills (HOTS) worksheet for Class 10 Probability. Advanced problems for CBSE Board Exams.

HOTS (High Order Thinking Skills)

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  1. Question 1: A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
    Solution:
    Step 1: Total marbles = 24. Let green marbles = $g$.
    Step 2: $P(\text{Green}) = \frac{g}{24} = \frac{2}{3} \Rightarrow g = 16$.
    Step 3: Blue marbles = Total - Green = $24 - 16 = 8$.
    Answer: 8 blue marbles.
  2. Question 2: A number $x$ is selected from the numbers 1, 2, 3 and then a second number $y$ is randomly selected from the numbers 1, 4, 9. What is the probability that the product $xy$ of the two numbers will be less than 9?
    Solution:
    Step 1: Possible pairs $(x, y)$: $(1,1), (1,4), (1,9), (2,1), (2,4), (2,9), (3,1), (3,4), (3,9)$. Total = 9.
    Step 2: Products: $1, 4, 9, 2, 8, 18, 3, 12, 27$.
    Step 3: Products less than 9: $1, 4, 2, 8, 3$. Count = 5.
    Step 4: Probability $= 5/9$.
    Answer: 5/9.
  3. Question 3: Find the probability that a leap year selected at random will contain 53 Sundays and 53 Mondays.
    Solution:
    Step 1: A leap year has 366 days = 52 weeks + 2 extra days.
    Step 2: Possible pairs for extra days: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Total = 7.
    Step 3: For 53 Sundays AND 53 Mondays, the extra days must be (Sun, Mon).
    Step 4: Favourable outcomes = 1. Probability $= 1/7$.
    Answer: 1/7.
  4. Question 4: A bag contains 12 balls out of which $x$ are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball? (ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (i). Find $x$.
    Solution:
    Step 1: (i) Total = 12, White = $x$. $P_1 = x/12$.
    Step 2: (ii) New Total = $12+6=18$, New White = $x+6$. $P_2 = \frac{x+6}{18}$.
    Step 3: Given $P_2 = 2P_1 \Rightarrow \frac{x+6}{18} = 2(\frac{x}{12}) \Rightarrow \frac{x+6}{18} = \frac{x}{6}$.
    Step 4: $x+6 = 3x \Rightarrow 2x = 6 \Rightarrow x = 3$.
    Answer: $x=3$.
  5. Question 5: All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar to other cards, find the probability that the card has a value (i) 7 (ii) greater than 7 (iii) less than 7.
    Solution:
    Step 1: Removed cards: $3 \times 4 = 12$. Remaining cards $= 52 - 12 = 40$. (Cards are A, 2, 3, ..., 10 of each suit).
    Step 2: (i) Value 7: 4 cards (one of each suit). Prob $= 4/40 = 1/10$.
    Step 3: (ii) Value > 7: 8, 9, 10 (3 values $\times$ 4 suits = 12 cards). Prob $= 12/40 = 3/10$.
    Step 4: (iii) Value < 7: A, 2, 3, 4, 5, 6 (6 values $\times$ 4 suits = 24 cards). Prob $= 24/40 = 3/5$.
    Answer: (i) 1/10, (ii) 3/10, (iii) 3/5.
  6. Question 6: The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3. If the jar contains 10 orange balls, find the total number of balls in the jar.
    Solution:
    Step 1: Let total balls be $N$. $P(R) = 1/4, P(B) = 1/3$.
    Step 2: $P(O) = 1 - (1/4 + 1/3) = 1 - 7/12 = 5/12$.
    Step 3: Number of orange balls = 10. So, $\frac{5}{12} N = 10$.
    Step 4: $N = \frac{10 \times 12}{5} = 24$.
    Answer: 24 balls.
  7. Question 7: Two dice are thrown at the same time. Find the probability of getting (i) same number on both dice (ii) different numbers on both dice.
    Solution:
    Step 1: Total outcomes = 36.
    Step 2: (i) Same numbers (doublets): (1,1), (2,2)...(6,6). Count = 6. Prob $= 6/36 = 1/6$.
    Step 3: (ii) Different numbers: Total - Same = $36 - 6 = 30$. Prob $= 30/36 = 5/6$.
    Answer: (i) 1/6, (ii) 5/6.
  8. Question 8: A box contains cards bearing numbers from 6 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a one digit number (ii) a number divisible by 5 (iii) an odd number less than 30.
    Solution:
    Step 1: Total cards $= 70 - 6 + 1 = 65$.
    Step 2: (i) One digit: 6, 7, 8, 9. Count = 4. Prob $= 4/65$.
    Step 3: (ii) Divisible by 5: 10, 15, ..., 70. AP: $a=10, l=70, d=5$. $70 = 10 + (n-1)5 \Rightarrow 60 = 5(n-1) \Rightarrow n=13$. Prob $= 13/65 = 1/5$.
    Step 4: (iii) Odd < 30: 7, 9, ..., 29. Count = 12. Prob $= 12/65$.
    Answer: (i) 4/65, (ii) 1/5, (iii) 12/65.
  9. Question 9: Three coins are tossed simultaneously. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at most two heads.
    Solution:
    Step 1: Total outcomes = 8 (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT).
    Step 2: (i) Exactly 2 heads: HHT, HTH, THH. Count = 3. Prob $= 3/8$.
    Step 3: (ii) At least 2 heads: HHH, HHT, HTH, THH. Count = 4. Prob $= 4/8 = 1/2$.
    Step 4: (iii) At most 2 heads: All except HHH. Count = 7. Prob $= 7/8$.
    Answer: (i) 3/8, (ii) 1/2, (iii) 7/8.
  10. Question 10: A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
    Solution:
    Step 1: Total outcomes = 36.
    Step 2: (ii) 5 comes up at least once: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6). Count = 11. Prob $= 11/36$.
    Step 3: (i) 5 will not come up = $1 - P(\text{at least once}) = 1 - 11/36 = 25/36$.
    Answer: (i) 25/36, (ii) 11/36.
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