Chapter 14 : Probability Class 10 Maths NCERT Solutions

Q11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

• The Logic: This is a straightforward “Random Selection” problem. We need to find the total number of fish in the tank and then identify how many of them meet Gopi’s “male fish” criteria.

• The Steps:

  1. Number of male fish = 5.

  2. Number of female fish = 8.

  3. Total number of fish = 5 + 8 = 13.

  4. P(male fish) = Favourable outcomes / Total outcomes = 5 / 13.

• The Result: The probability that the fish is a male fish is 5/13.

Q12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

• The Logic: The spinner has 8 equal sections. Since the outcomes are “equally likely,” the denominator for every part of this question will be 8.

• The Steps:

  1. Total outcomes = {1, 2, 3, 4, 5, 6, 7, 8}. Total = 8.

  2. (i) Pointing at 8: Favourable outcome is {8}. Count = 1.

     • P(8) = 1/8.

  3. (ii) An odd number: Favourable outcomes are {1, 3, 5, 7}. Count = 4.

     • P(odd) = 4/8 = 1/2.

  4. (iii) A number greater than 2: Favourable outcomes are {3, 4, 5, 6, 7, 8}. Count = 6.

     • P(> 2) = 6/8 = 3/4.

  5. (iv) A number less than 9: All numbers {1, 2, 3, 4, 5, 6, 7, 8} are less than 9. Count = 8.

     • P(< 9) = 8/8 = 1.

• The Result: (i) 1/8, (ii) 1/2, (iii) 3/4, (iv) 1

Q13. A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

• The Logic: A single die has six faces. Human Tip: Students often forget that ‘1’ is neither prime nor composite. The prime numbers on a die are 2, 3, and 5.

• The Steps:

  1. Total outcomes = {1, 2, 3, 4, 5, 6}. Total = 6.

  2. (i) A prime number: Favourable outcomes are {2, 3, 5}. Count = 3.

     • P(prime) = 3/6 = 1/2.

  3. (ii) Between 2 and 6: Favourable outcomes are {3, 4, 5}. Count = 3.

     • P(between 2 and 6) = 3/6 = 1/2.

  4. (iii) An odd number: Favourable outcomes are {1, 3, 5}. Count = 3.

     • P(odd) = 3/6 = 1/2.

• The Result: (i) 1/2, (ii) 1/2, (iii) 1/2

Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.

• The Logic: To solve card problems, you need to know the deck structure. There are 4 suits (Spades, Hearts, Diamonds, Clubs), each with 13 cards. Face cards are Jacks, Queens, and Kings (3 per suit).

• The Steps:

  1. Total outcomes = 52.

  2. (i) King of red colour: Red suits are Hearts and Diamonds. Each has one King. Favourable = 2.

     • P = 2/52 = 1/26.

  3. (ii) A face card: 3 face cards per suit * 4 suits = 12.

     • P = 12/52 = 3/13.

  4. (iii) A red face card: 3 face cards in Hearts + 3 in Diamonds = 6.

     • P = 6/52 = 3/26.

  5. (iv) The jack of hearts: Only 1 exists.

     • P = 1/52.

  6. (v) A spade: There are 13 cards in the spade suit.

     • P = 13/52 = 1/4.

  7. (vi) The queen of diamonds: Only 1 exists.

     • P = 1/52.

• The Result: (i) 1/26, (ii) 3/13, (iii) 3/26, (iv) 1/52, (v) 1/4, (vi) 1/52

Q15. Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

• Pro-Tip: Pay close attention to the phrase “put aside.” This changes your total number of cards (the denominator) for the second pick!

• The Steps:

   

  1. Initial Set: {10, Jack, Queen, King, Ace} of Diamonds. Total = 5.

  2. (i) Picking a Queen: Favourable = 1.

     • P(Queen) = 1/5.

  3. (ii) Queen is removed: New set = {10, Jack, King, Ace}. New Total = 4.

     • (a) Picking an Ace: Favourable = 1. P = 1/4.

     • (b) Picking a Queen: There are 0 queens left. P = 0/4 = 0

• The Result: (i) 1/5, (ii) (a) 1/4, (b) 0

Q16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this mixture. Determine the probability that the pen taken out is a good one.

• The Logic: This is a “total pool” problem. To find the probability of picking a good pen, you first need to know the total number of pens in the mixture.

• The Steps:

  1. Number of defective pens = 12.

  2. Number of good pens = 132.

  3. Total number of pens = 12 + 132 = 144.

  4. P(good pen) = Favourable outcomes / Total outcomes = 132 / 144.

  5. Simplify the fraction: Divide both by 12.

• The Result: The probability that the pen is a good one is 11/12.

Q17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

• The Logic: Part (ii) is a “Conditional Probability” scenario. When you remove a good bulb and don’t put it back, both the total number of bulbs and the number of good bulbs decrease by 1.

• The Steps:

  1. (i) First Draw: Total bulbs = 20. Defective = 4.

     • P(defective) = 4 / 20 = 1 / 5.

  2. (ii) Second Draw: A “not defective” (good) bulb was removed.

     • New total bulbs = 20 – 1 = 19.

     • Original good bulbs = 20 – 4 = 16.

     • New good bulbs = 16 – 1 = 15.

     • P(not defective) = 15 / 19.

• The Result: (i) 1/5, (ii) 15/19

Q18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

• The Logic: Your total outcomes are 90. For each part, you need to carefully count how many numbers between 1 and 90 fit the criteria.

• The Steps:

  1. Total outcomes = 90.

  2. (i) Two-digit numbers: Numbers from 10 to 90. (Total numbers – one-digit numbers {1 to 9}) = 90 – 9 = 81.

     • P = 81 / 90 = 9 / 10.

  3. (ii) Perfect square numbers: {1, 4, 9, 16, 25, 36, 49, 64, 81}. (Note: 10*10 = 100, which is too high). Total = 9.

     • P = 9 / 90 = 1 / 10.

  4. (iii) Divisible by 5: {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}. Total = 18.

     • P = 18 / 90 = 1 / 5.

• The Result: (i) 9/10, (ii) 1/10, (iii) 1/5

Q19. A child has a die whose six faces show the letters as given below: A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?

• The Logic: This is just like a standard die, but the “favourable” outcomes are letters instead of numbers. Notice that the letter ‘A’ appears twice.

• The Steps:

  1. Total outcomes = {A, B, C, D, E, A}. Total = 6.

  2. (i) Getting A: Favourable outcomes = 2 (since there are two ‘A’s).

     • P(A) = 2 / 6 = 1 / 3.

  3. (ii) Getting D: Favourable outcomes = 1.

     • P(D) = 1 / 6.

• The Result: (i) 1/3, (ii) 1/6

Q20. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?

• The Logic: This is Geometric Probability. Instead of counting items, we use areas. The probability is the (Area of the target region / Area of the total region).

• The Steps:

  1. Total Area (Rectangle): Length * Breadth = 3 m * 2 m = 6 m^2.

  2. Target Area (Circle): Diameter = 1 m, so Radius (r) = 1/2 m.

     • Area of circle = pi * r^2 = pi * (1/2)^2 = pi / 4 m^2.

  3. Probability: (Area of Circle) / (Area of Rectangle)

     • P = (pi / 4) / 6 = pi / 24.

• The Result: The probability is pi / 24.

Q21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?

• The Logic: Nuri’s decision depends entirely on the quality of the pen. To find the probability, we first separate the “good” from the “bad” and then compare them to the total pool of 144 pens.

• The Steps:

  1. Total number of pens = 144.

  2. Number of defective pens = 20.

  3. Number of good pens = 144 – 20 = 124.

  4. (i) Probability she will buy it: This is the probability of picking a good pen.

     • P(Buy) = 124 / 144.

     • Simplify (divide by 4): 31 / 36.

  5. (ii) Probability she will not buy it: This is the probability of picking a defective pen.

     • P(Not Buy) = 20 / 144.

     • Simplify (divide by 4): 5 / 36.

• The Result: (i) 31/36, (ii) 5/36

Q22. Refer to Example 13. (i) Complete the following table:

(Table: Sum on 2 dice: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11.’ Do you agree with this argument? Justify your answer.

• The Logic: When two dice are thrown, there are 6 * 6 = 36 total outcomes. While there are 11 possible “sums,” they are not equally likely because some sums can be formed in many ways (like 7), while others only in one way (like 2 or 12).

• The Steps:

  1. Total outcomes = 36.

  2. (i) Completed Table (Probability of Sums):

     • Sum 2: (1,1) -> 1/36

     • Sum 3: (1,2), (2,1) -> 2/36

     • Sum 4: (1,3), (2,2), (3,1) -> 3/36

     • Sum 5: (1,4), (2,3), (3,2), (4,1) -> 4/36

     • Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) -> 5/36

     • Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) -> 6/36

     • Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) -> 5/36

     • Sum 9: (3,6), (4,5), (5,4), (6,3) -> 4/36

     • Sum 10: (4,6), (5,5), (6,4) -> 3/36

     • Sum 11: (5,6), (6,5) -> 2/36

     • Sum 12: (6,6) -> 1/36

  3. (ii) Justification: No, the argument is incorrect. The 11 outcomes (sums) are not equally likely. As seen in the table, the probabilities range from 1/36 to 6/36.

• The Result: (i) Table completed as above. (ii) No, the argument is incorrect.

Q23. A game consists of tossing a one 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.

• The Logic: In a triple coin toss, there are 8 possible combinations. We find the “winning” ones first, and the rest are Hanif’s losses.

   

• The Steps:

  1. Total Outcomes: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total = 8.

  2. Winning Cases (Same result): {HHH, TTT}. Count = 2.

  3. Losing Cases: All other outcomes = 8 – 2 = 6.

  4. P(Losing) = Favourable outcomes / Total outcomes = 6 / 8.

  5. Simplify (divide by 2): 3 / 4.

• The Result: The probability that Hanif will lose is 3/4 (or 0.75).

Q24. 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? [Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

• The Logic: Throwing a die twice gives 36 outcomes. It is easier to find where 5 does appear first (at least once) and then subtract that from the total to find where it doesn’t appear.

   

• The Steps:

  1. Total Outcomes = 36.

  2. Cases where 5 comes up at least once:

     • First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) -> 6 cases.

     • Second die is 5: (1,5), (2,5), (3,5), (4,5), (6,5) -> 5 cases (we don’t count 5,5 twice!).

     • Total favourable for (ii) = 6 + 5 = 11.

  3. (ii) P(5 at least once) = 11 / 36.

  4. (i) P(5 not either time): Total cases – Cases where 5 appears = 36 – 11 = 25.

     • P(No 5) = 25 / 36.

• The Result: (i) 25/36, (ii) 11/36.

Q25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.

(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

• The Logic: We check if the outcomes are “equally likely.” If they aren’t, the 1/n probability rule doesn’t apply.

   

• The Steps:

  1. Argument (i): Incorrect. When two coins are tossed, the outcomes are {HH, HT, TH, TT}. “One of each” (HT and TH) happens twice as often as “two heads.” The true probabilities are 1/4 (HH), 1/4 (TT), and 2/4 = 1/2 (one of each).

  2. Argument (ii): Correct. A die has {1, 2, 3, 4, 5, 6}. Odd numbers are {1, 3, 5} and even numbers are {2, 4, 6}. Both groups have 3 outcomes out of 6, making the probability 3/6 = 1/2 for each.

• The Result: (i) Incorrect, (ii) Correct.