CBSE Class 10 Probability Sure Shot Questions Set 01

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PRACTICE QUESTIONS
CLASS X : CHAPTER - 15
PROBABILITY : DICE BASED QUESTIONS

1. An unbiased die is thrown. What is the probability of getting
a). an even number
b). a multiple of 3
c). a multiple of 2 or 3
d). a number less than 5 divisible by 2.
e). A number greater than 2 divisible by 3.
f). an even number or a multiple of 3
g). an even number and a multiple of 3
h). a number 3 or 4
i). an odd number
j). a number less than 5
k). a number greater than 3
l). a number between 3 and 6. [3 Marks]

Answer:
1. Total possible outcomes when a die is thrown = 6 ({1, 2, 3, 4, 5, 6}).
2. (a) Even numbers are 2, 4, 6 (Favorable outcomes = 3); Probability = \(\frac{3}{6}\) = \(\frac{1}{2}\).
3. (b) Multiples of 3 are 3, 6 (Favorable outcomes = 2); Probability = \(\frac{2}{6}\) = \(\frac{1}{3}\).
4. (c) Multiples of 2 or 3 are 2, 3, 4, 6 (Favorable outcomes = 4); Probability = \(\frac{4}{6}\) = \(\frac{2}{3}\).
5. (d) Less than 5 and divisible by 2 are 2, 4 (Favorable outcomes = 2); Probability = \(\frac{2}{6}\) = \(\frac{1}{3}\).
6. (e) Greater than 2 and divisible by 3 is 3, 6 (Favorable outcomes = 2); Probability = \(\frac{2}{6}\) = \(\frac{1}{3}\).
7. (f) Even or multiple of 3 are 2, 3, 4, 6 (Favorable outcomes = 4); Probability = \(\frac{4}{6}\) = \(\frac{2}{3}\).
8. (g) Even and multiple of 3 is 6 (Favorable outcomes = 1); Probability = \(\frac{1}{6}\).
9. (h) Numbers 3 or 4 are 3, 4 (Favorable outcomes = 2); Probability = \(\frac{2}{6}\) = \(\frac{1}{3}\).
10. (i) Odd numbers are 1, 3, 5 (Favorable outcomes = 3); Probability = \(\frac{3}{6}\) = \(\frac{1}{2}\).
11. (j) Numbers less than 5 are 1, 2, 3, 4 (Favorable outcomes = 4); Probability = \(\frac{4}{6}\) = \(\frac{2}{3}\).
12. (k) Numbers greater than 3 are 4, 5, 6 (Favorable outcomes = 3); Probability = \(\frac{3}{6}\) = \(\frac{1}{2}\).
13. (l) Numbers between 3 and 6 are 4, 5 (Favorable outcomes = 2); Probability = \(\frac{2}{6}\) = \(\frac{1}{3}\).

Teacher's Note:
a) Probability is defined as the ratio of favorable outcomes to the total number of equally likely outcomes.
b) For compound conditions like 'or', take the union of sets, and for 'and', take the intersection of sets of outcomes.

2. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is (i) 6 (ii) 12 (iii) 7 [3 Marks]

Answer:
1. Total possible outcomes when two dice are thrown = 6 × 6 = 36.
2. (i) Product is 6: favorable pairs are (1, 6), (2, 3), (3, 2), (6, 1). Number of outcomes = 4. Probability = \(\frac{4}{36}\) = \(\frac{1}{9}\).
3. (ii) Product is 12: favorable pairs are (2, 6), (3, 4), (4, 3), (6, 2). Number of outcomes = 4. Probability = \(\frac{4}{36}\) = \(\frac{1}{9}\).
4. (iii) Product is 7: no two integers between 1 and 6 have a product of 7. Number of outcomes = 0. Probability = 0.

Teacher's Note:
a) List all possible ordered pairs systematically to avoid missing any combination.
b) A product of a prime number like 7 is impossible when choosing integers from 1 to 6.

3. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9. [3 Marks]

Answer:
1. Total possible outcomes = 36.
2. List outcomes where the product of the two numbers is strictly less than 9:
- Product 1: (1, 1) [1 outcome]
- Product 2: (1, 2), (2, 1) [2 outcomes]
- Product 3: (1, 3), (3, 1) [2 outcomes]
- Product 4: (1, 4), (2, 2), (4, 1) [3 outcomes]
- Product 5: (1, 5), (5, 1) [2 outcomes]
- Product 6: (1, 6), (2, 3), (3, 2), (6, 1) [4 outcomes]
- Product 7: none [0 outcomes]
- Product 8: (2, 4), (4, 2) [2 outcomes]
3. Total favorable outcomes = 1 + 2 + 2 + 3 + 2 + 4 + 0 + 2 = 16.
4. Probability = \(\frac{16}{36}\) = \(\frac{4}{9}\).

Teacher's Note:
a) Carefully count products row by row or column by column using a 6 by 6 grid matrix.
b) Ensure strict inequality (less than 9) is maintained so products equal to 9 are excluded.

4. Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately. [4 Marks]

Answer:
1. Total possible outcomes = 6 × 6 = 36.
2. The first die has faces {1, 2, 3, 4, 5, 6} and the second die has faces {1, 1, 2, 2, 3, 3}.
3. Sum = 2: outcomes (1, 1), (1, 1) -> 2 outcomes. Probability = \(\frac{2}{36}\) = \(\frac{1}{18}\).
4. Sum = 3: outcomes (1, 2), (1, 2), (2, 1), (2, 1) -> 4 outcomes. Probability = \(\frac{4}{36}\) = \(\frac{1}{9}\).
5. Sum = 4: outcomes (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1) -> 6 outcomes. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
6. Sum = 5: outcomes (2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1) -> 6 outcomes. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
7. Sum = 6: outcomes (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1) -> 6 outcomes. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
8. Sum = 7: outcomes (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1) -> 6 outcomes. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
9. Sum = 8: outcomes (5, 3), (5, 3), (6, 2), (6, 2) -> 4 outcomes. Probability = \(\frac{4}{36}\) = \(\frac{1}{9}\).
10. Sum = 9: outcomes (6, 3), (6, 3) -> 2 outcomes. Probability = \(\frac{2}{36}\) = \(\frac{1}{18}\).

Teacher's Note:
a) Recognize that the second die has repeated values, changing the frequency of each outcome compared to standard dice.
b) Verify that the sum of all probabilities equals 1: \(\frac{2+4+6+6+6+6+4+2}{36} = \frac{36}{36} = 1\).

5. Two dice are thrown at the same time. Determine the probabiity that the difference of the numbers on the two dice is 2. [2 Marks]

Answer:
1. Total possible outcomes = 36.
2. Favorable outcomes where the absolute difference between the two numbers is 2 are:
- (1, 3), (3, 1)
- (2, 4), (4, 2)
- (3, 5), (5, 3)
- (4, 6), (6, 4)
3. Total favorable outcomes = 8.
4. Probability = \(\frac{8}{36}\) = \(\frac{2}{9}\).

Teacher's Note:
a) Difference means absolute difference, so both (x, y) and (y, x) must be considered when x != y.
b) Systematic listing by fixed differences helps prevent counting errors.

6. 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. [2 Marks]

Answer:
1. Total possible outcomes = 36.
2. (i) Same number on both dice: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). Total outcomes = 6.
Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
3. (ii) Different numbers on both dice: Complement of getting the same number.
Probability = 1 - \(\frac{1}{6}\) = \(\frac{5}{6}\).

Teacher's Note:
a) Utilize the complementary event rule to find the probability of different numbers quickly.
b) Total outcomes always decompose into mutually exclusive and exhaustive sub-events.

7. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is (i) 7? (ii) a prime number? (iii) 1? [3 Marks]

Answer:
1. Total possible outcomes = 36.
2. (i) Sum is 7: favorable outcomes are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Total = 6.
Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
3. (ii) Sum is a prime number (possible sums are 2, 3, 5, 7, 11):
- Sum 2: (1, 1) [1]
- Sum 3: (1, 2), (2, 1) [2]
- Sum 5: (1, 4), (2, 3), (3, 2), (4, 1) [4]
- Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) [6]
- Sum 11: (5, 6), (6, 5) [2]
Total favorable outcomes = 1 + 2 + 4 + 6 + 2 = 15.
Probability = \(\frac{15}{36}\) = \(\frac{5}{12}\).
4. (iii) Sum is 1: minimum possible sum with two dice is 1 + 1 = 2, so sum of 1 is impossible. Probability = 0.

Teacher's Note:
a) Recall that prime numbers greater than 1 are divisible only by 1 and themselves (2, 3, 5, 7, 11 are the possible sums here).
b) The minimum sum of two standard dice is 2, making probability for sum 1 identically zero.

8. Two dice are thrown simultaneously. Find the probability of getting
a). an even number on first dice
b). an odd number on first dice
c). an even number as the sum
d). a multiple of 5 as the sum
e). a multiple of 7 as the sum
f). a multiple of 3 as the sum
g). a sum more than 7
h). a sum greater than 9
i). neither the sum 9 nor the sum 11 as the sum
j). a sum less than 6 [4 Marks]

Answer:
1. Total possible outcomes = 36.
2. (a) Even number on first die (2, 4, 6 for first coordinate): 3 × 6 = 18 outcomes. Probability = \(\frac{18}{36}\) = \(\frac{1}{2}\).
3. (b) Odd number on first die (1, 3, 5 for first coordinate): 3 × 6 = 18 outcomes. Probability = \(\frac{18}{36}\) = \(\frac{1}{2}\).
4. (c) Even number as the sum (sums 2, 4, 6, 8, 10, 12): 1 + 3 + 5 + 5 + 3 + 1 = 18 outcomes. Probability = \(\frac{18}{36}\) = \(\frac{1}{2}\).
5. (d) Multiple of 5 as the sum (sums 5, 10):
- Sum 5: (1, 4), (2, 3), (3, 2), (4, 1) [4]
- Sum 10: (4, 6), (5, 5), (6, 4) [3]
Total = 7 outcomes. Probability = \(\frac{7}{36}\).
6. (e) Multiple of 7 as the sum (sum 7): 6 outcomes. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
7. (f) Multiple of 3 as the sum (sums 3, 6, 9, 12):
- Sum 3: 2
- Sum 6: 5
- Sum 9: 4
- Sum 12: 1
Total = 2 + 5 + 4 + 1 = 12 outcomes. Probability = \(\frac{12}{36}\) = \(\frac{1}{3}\).
8. (g) Sum more than 7 (sums 8, 9, 10, 11, 12): 5 + 4 + 3 + 2 + 1 = 15 outcomes. Probability = \(\frac{15}{36}\) = \(\frac{5}{12}\).
9. (h) Sum greater than 9 (sums 10, 11, 12): 3 + 2 + 1 = 6 outcomes. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).
10. (i) Neither sum 9 nor sum 11: Total outcomes for sums 9 and 11 combined is 4 + 2 = 6. Favorable outcomes = 36 - 6 = 30. Probability = \(\frac{30}{36}\) = \(\frac{5}{6}\).
11. (j) Sum less than 6 (sums 2, 3, 4, 5): 1 + 2 + 3 + 4 = 10 outcomes. Probability = \(\frac{10}{36}\) = \(\frac{5}{18}\).

Teacher's Note:
a) Break down complex sum queries by listing individual outcomes per sum value.
b) Use symmetry properties of dice sums: frequency distribution of sums from 2 to 12 is symmetric around 7.
c) Apply the complement rule efficiently for negative constraints like 'neither'.
d) Double check arithmetic sums of frequencies to ensure they total 36.

k). Find the probability of getting a sum less than 7 when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes when two dice are thrown = \(6 \times 6 = 36\).
2. Favorable outcomes for a sum less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1), total 15 outcomes.
3. Probability = \(\frac{15}{36}\) = \(\frac{5}{12}\).

Teacher's Note:
a) The sum less than 7 excludes sums equal to 7.
b) List outcomes systematically row by row to avoid missing any pair.

l). Find the probability of getting a sum more than 7 when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Favorable outcomes for a sum more than 7 are those with sums 8, 9, 10, 11, and 12, totaling 15 outcomes.
3. Probability = \(\frac{15}{36}\) = \(\frac{5}{12}\).

Teacher's Note:
a) Sums greater than 7 start from 8 up to the maximum possible sum of 12.
b) Use the complementary probability method as an alternative check.

m). Find the probability of getting a multiple of 3 on one dice when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Multiples of 3 on a single die are 3 and 6.
3. Favorable outcomes include pairs where at least one die shows 3 or 6, which are 20 outcomes.
4. Probability = \(\frac{20}{36}\) = \(\frac{5}{9}\).

Teacher's Note:
a) At least one die means either the first, the second, or both.
b) Subtract the outcomes where neither die is a multiple of 3 from the total.

n). Find the probability of getting a multiple of 2 on one dice when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Multiples of 2 on a single die are 2, 4, and 6.
3. Favorable outcomes where at least one die shows a multiple of 2 total 27 outcomes.
4. Probability = \(\frac{27}{36}\) = \(\frac{3}{4}\).

Teacher's Note:
a) Multiples of 2 are even numbers.
b) Use complementary counting by finding outcomes where both dice show odd numbers.

o). Find the probability of getting a multiple of 5 on one dice when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. The only multiple of 5 on a standard die is 5.
3. Favorable outcomes having at least one 5 are 11 outcomes.
4. Probability = \(\frac{11}{36}\).

Teacher's Note:
a) Be careful not to double count the outcome (5,5).
b) Count outcomes with exactly one 5 and those with two 5s separately.

p). Find the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Multiples of 2 on the first die are {2, 4, 6} and multiples of 3 on the second die are {3, 6}, giving 3 × 2 = 6 outcomes.
3. Multiples of 3 on the first die are {3, 6} and multiples of 2 on the second die are {2, 4, 6}, giving 2 × 3 = 6 outcomes.
4. Total favorable outcomes considering overlap is 10.
5. Probability = \(\frac{10}{36}\) = \(\frac{5}{18}\).

Teacher's Note:
a) Order matters as the condition specifies roles for each die.
b) Check for intersection where both conditions overlap.

q). Find the probability of getting a doublet when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Favorable outcomes are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6), totaling 6 outcomes.
3. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).

Teacher's Note:
a) A doublet means both dice show the exact same number.
b) There are 6 possible doublets out of 36 sample points.

r). Find the probability of getting a doublet of even number when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Even number doublets are (2,2), (4,4), and (6,6), totaling 3 outcomes.
3. Probability = \(\frac{3}{36}\) = \(\frac{1}{12}\).

Teacher's Note:
a) Even numbers on a die are 2, 4, and 6.
b) Restrict the doublet set to only even entries.

s). Find the probability of getting a doublet of odd number when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Odd number doublets are (1,1), (3,3), and (5,5), totaling 3 outcomes.
3. Probability = \(\frac{3}{36}\) = \(\frac{1}{12}\).

Teacher's Note:
a) Odd numbers on a die are 1, 3, and 5.
b) Symmetry exists between odd and even doublets.

t). Find the probability of getting a doublet of prime number when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Prime numbers on a die are 2, 3, and 5.
3. Prime doublets are (2,2), (3,3), and (5,5), totaling 3 outcomes.
4. Probability = \(\frac{3}{36}\) = \(\frac{1}{12}\).

Teacher's Note:
a) Note that 1 is neither prime nor composite.
b) Prime doublets are specifically formed by 2, 3, and 5.

u). Find the probability of getting a number other than 5 on any dice when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Outcomes where 5 does not appear on either die are \(5 \times 5 = 25\).
3. Probability = \(\frac{25}{36}\).

Teacher's Note:
a) "Other than 5" means neither die shows a 5.
b) Using the complement of getting at least one 5 confirms 1 - 11/36 = 25/36.

v). Find the probability of getting a number other than 3 on any dice when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Outcomes where 3 does not appear on either die are \(5 \times 5 = 25\).
3. Probability = \(\frac{25}{36}\).

Teacher's Note:
a) Exclude all pairs containing the number 3.
b) Both dice must independently avoid showing a 3.

w). Find the probability of getting the sum equal to 12 when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. The only outcome that yields a sum of 12 is (6,6), totaling 1 outcome.
3. Probability = \(\frac{1}{36}\).

Teacher's Note:
a) A sum of 12 is the maximum possible sum for two standard dice.
b) Only a single combination satisfies this condition.

x). Find the probability of getting the sum greater than equal to 10 when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Favorable outcomes for sums greater than or equal to 10 are sums of 10, 11, and 12.
3. These pairs are (4,6), (5,5), (6,4), (5,6), (6,5), and (6,6), totaling 6 outcomes.
4. Probability = \(\frac{6}{36}\) = \(\frac{1}{6}\).

Teacher's Note:
a) "Greater than equal to" includes the boundary value 10.
b) Count pairs summing to 10, 11, and 12 carefully.

y). Find the probability of getting the sum less than or equal to 10 when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Outcomes with a sum greater than 10 are (5,6), (6,5), and (6,6), totaling 3 outcomes.
3. Favorable outcomes for sum less than or equal to 10 = 36 - 3 = 33.
4. Probability = \(\frac{33}{36}\) = \(\frac{11}{12}\).

Teacher's Note:
a) Using the complementary event makes calculation faster.
b) Complement of sum less than or equal to 10 is sum greater than 10.

z). Find the probability of getting the sum as a prime number when two dice are thrown together. [1 Mark]

Answer:
1. Total possible outcomes = 36.
2. Possible sums when two dice are thrown range from 2 to 12.
3. Prime sums in this range are 2, 3, 5, 7, and 11.
4. Favorable outcomes for each prime sum:
    - Sum 2: (1,1) [1]
    - Sum 3: (1,2), (2,1) [2]
    - Sum 5: (1,4), (2,3), (3,2), (4,1) [4]
    - Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) [6]
    - Sum 11: (5,6), (6,5) [2]
5. Total favorable outcomes = 1 + 2 + 4 + 6 + 2 = 15.
6. Probability = \(\frac{15}{36}\) = \(\frac{5}{12}\).

Teacher's Note:
a) Prime numbers are integers greater than 1 with no positive divisors other than 1 and itself.
b) Grouping by possible prime sums ensures no combination is missed or counted twice.

Multiple Choice Questions

Question. The probability of getting exactly one head in tossing a pair of coins is
(a) 0
(b) 1
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{2} \)
Answer: (d)
Explanation: \( S = [HH, HT, TH, TT] = 4 \)
\( \therefore P(\text{exactly 1 head}) = \frac{2}{4} = \frac{1}{2} \)

Question. The probability of getting a spade card from a well shuffled deck of 52 cards is
(a) \( \frac{1}{13} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{12}{13} \)
(d) \( \frac{3}{4} \)
Answer: (b)
Explanation: Total cards = 52, Spade cards = 13
\( \therefore P(\text{a spade card}) = \frac{13}{52} = \frac{1}{4} \)

Question. The probability of getting less than 3 in a single throw of a die is
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{2}{3} \)
Answer: (a)
Explanation: Here \( S = 6^1 = 6 \) and \( E = (\text{less than 3}) = [1, 2] = 2 \)
\( \therefore P(\text{Less than 3}) = \frac{2}{6} = \frac{1}{3} \)

Question. The total number of events of throwing 10 coins simultaneously is
(a) 1024
(b) 512
(c) 100
(d) 10
Answer: (a)
Explanation: Total events = \( 2^{10} = 1024 \)

Question. Which of the following can be the probability of an event?
(a) – 0.4
(b) 1.004
(c) \( \frac{18}{23} \)
(d) \( \frac{10}{7} \)
Answer: (c)
Explanation: Event can neither be a -ve, nor it can be \( > 1 \).

Question. Three coins are tossed simultaneously. The probability of getting all heads is
(a) 1
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{8} \)
Answer: (d)
Explanation: Here \( S = [HHH, HHT, HTH, THH, HTT, THT, TTH, TTT] = 8 \)
\( \therefore P(\text{all heads}) = \frac{1}{8} \)

Question. One card is drawn from a well shuffled deck of 52 cards. The probability of getting a king of red colour is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{2} \)
Answer: (a)
Explanation: Total cards = 52
\( \therefore \) Total events \( n(S) = 52 \) and A king of red colour = 2
\( \therefore P(\text{a king of red colour}) = \frac{2}{52} = \frac{1}{26} \)

Question. One card is drawn from a well shuffled deck of 52 playing cards. The probability of getting a non-face card is
(a) \( \frac{3}{13} \)
(b) \( \frac{10}{13} \)
(c) \( \frac{7}{13} \)
(d) \( \frac{4}{13} \)
Answer: (b)
Explanation: Total cards = 52, Total Face cards = 12
\( \therefore \) Non-face cards = \( 52 - 12 = 40 \)
\( \therefore P(\text{a non-face card}) = \frac{40}{52} = \frac{10}{13} \)

Question. The chance of throwing 5 with an ordinary die is
(a) \( \frac{1}{6} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{2} \)
Answer: (a)
Explanation: Here \( S = [1, 2, 3, 4, 5, 6] \therefore n(S) = 6 \)
\( \therefore P(\text{throwings}) = \frac{1}{6} \)

Question. The letters of the word SOCIETY are placed at random in a row. The probability of getting a vowel is
(a) \( \frac{1}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{3}{7} \)
(d) \( \frac{4}{7} \)
Answer: (c)
Explanation: Total letters = 7; No. of vowels = O, I, E = 3
\( \therefore P(\text{a vowel}) = \frac{3}{7} \)

Question. Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out from the bag at random. The probability that the number on the card taken out is an even number, is
(a) \( \frac{1}{20} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{2} \)
Answer: (d)
Explanation: Total cards = 18
Cards with even no. are 4, 6, 8, 10, 12, 14, 16, 18, 20 = 9
\( \therefore P(\text{even number}) = \frac{9}{18} = \frac{1}{2} \)

Question. The total events to throw three dice simultaneously is
(a) 6
(b) 18
(c) 81
(d) 216
Answer: (d)
Explanation: Total cards = \( (6)^3 = 216 \)

Question. The probability of getting a consonant from the word MAHIR is
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{4}{5} \)
(d) 1
Answer: (b)
Explanation: Total characters in MAHIR = 5
Consonants are M, H, R i.e., 3
\( \therefore P(\text{getting a consonant}) = \frac{3}{5} \)

Question. A girl calculates that the probability of her winning the first prize in a lottery is 8/100. If 6,000 tickets are sold, how many tickets has she bought?
(a) 400
(b) 750
(c) 480
(d) 240
Answer: (c)
Explanation: No. of tickets sold = \( \frac{8}{100} \times 6000 = 480 \)

Question. A card is drawn from a well shuffled deck of 52 cards. The probability of a seven of spade is
(a) \( \frac{1}{26} \)
(b) \( \frac{1}{52} \)
(c) \( \frac{3}{52} \)
(d) \( \frac{1}{13} \)
Answer: (b)
Explanation: Total cards = 52, A seven of spade = 1
\( \therefore P(\text{a seven of spade}) = \frac{1}{52} \)

Question. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. The probability that a red ball drawn is
(a) \( \frac{3}{8} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{5}{8} \)
(d) \( \frac{3}{4} \)
Answer: (a)
Explanation: Total balls = 3 + 5 = 8
Here, Total events = 8 \( \therefore P(\text{a red ball}) = \frac{3}{8} \)

Question. A child has a die whose six faces show the letters as given below. The die is thrown once.
A B C D E F
The probability of getting a ‘D’ is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{6} \)
Answer: (d)
Explanation: Sample Space, \( S = [A, B, C, D, E, F] = 6 \)
\( \therefore n(S) = 6 \therefore P(\text{getting D}) = \frac{1}{6} \)

Question. One card is drawn from a well-shuffled deck of 52 cards. The probability that the card will not be an ace is
(a) \( \frac{1}{13} \)
(b) \( \frac{4}{13} \)
(c) \( \frac{12}{13} \)
(d) \( \frac{3}{13} \)
Answer: (c)
Explanation: Total cards = 52; No. of ace cards = 4
\( \therefore \) Non-ace cards = \( 52 - 4 = 48 \)
\( \therefore P(\text{not an ace}) = \frac{48}{52} = \frac{12}{13} \)

Question. A lot consists of 144 ball pens of which 20 ae defective and the others are good. Tanu will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. The probability that she will buy that pen is
(a) \( \frac{5}{36} \)
(b) \( \frac{20}{36} \)
(c) \( \frac{31}{36} \)
(d) \( \frac{31}{144} \)
Answer: (c)
Explanation: Total pens = 144; Defective pens = 20
Now, Good ball pens = \( 144 - 20 = 124 \)
\( \therefore P(\text{she will buy a pen}) = P(\text{good ball pen}) = \frac{124}{144} = \frac{31}{36} \)

Question. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(a) \( \frac{3}{5} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{4}{5} \)
Answer: (b)
Explanation: Total number = 40
No. of favourable events are 5, 10, 15, 20, 25, 30, 35, 40 = 8
\( \therefore P(\text{multiple of 5}) = \frac{8}{40} = \frac{1}{5} \)

Assertion-Reason Questions

Question. Assertion: The probability of winning a game is 0.4, then the probability of losing it, is 0.6.
Reason: \( P(E) + P(\text{not E}) = 1 \)
(a) Both Assertion (A) & Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.
Answer: (a)
Explanation: Given. \( P(E) = 0.4 \) ...where [E = event of winning]
\( \therefore P(\text{Not E}) = 1 - P(E) = 1 - 0.4 = 0.6 \)

Question. Assertion: The probability of getting a prime number when a die is thrown once is 2/3.
Reason: Prime numbers on a die are 2, 3, 5.
(a) Both Assertion (A) & Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.
Answer: (d)
Explanation: Given. Total outcomes = 6 and prime numbers = {2, 3, 5} = 3.
\( \therefore P(\text{Prime number}) = \frac{3}{6} = \frac{1}{2} \)

Question. Assertion: If a die is thrown, the probability of getting a number less than 3 & greater than 2 is zero.
Reason: Probability of an impossible event is zero.
(a) Both Assertion (A) & Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.
Answer: (a)
Explanation: Both statements are correct. Event given in Assertion is an impossible event.

Question. Assertion: Card numbered as 1, 2, 3 ........... 15 are put in a box and mixed thoroughly, one card is then drawn at random. The probability of drawing an even number is 7/15.
Reason: For any event E, we have \( 0 \le P(E) \le 1 \).
(a) Both Assertion (A) & Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.
Answer: (c)
Explanation: For any event E, we have, \( 0 \le P(E) \le 1 \). Even numbers between 1 and 15 are 2, 4, 6, 8, 10, 12, 14 which are 7. So probability is 7/15.

Question. Assertion: If a box contains 5 white, 2 red and 4 black marbles, then the probability of not drawing a white marble from the box is 5/11.
Reason: \( P(\bar{E}) = 1 – P(E) \), where E is any event.
(a) Both Assertion (A) & Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.
Answer: (d)
Explanation:
\( P(\text{white marble}) = \frac{\text{Possible outcomes}}{\text{Total outcomes}} = \frac{5}{5 + 2 + 4} = \frac{5}{11} \)
\( \therefore P(\text{not white marble}) = 1 - \frac{5}{11} = \frac{11 - 5}{11} = \frac{6}{11} \)

Case Based Questions

I. Rohit wants to distribute chocolates in his class on his birthday. The chocolates are of three types: Milk chocolate, White chocolate and Dark chocolate. If the total number of students in the class is 54 and everyone gets a chocolate, then answer the following questions:

Question. If the probability of distributing milk chocolates is \( \frac{1}{3} \), then the number of milk chocolates Rohit has, is
(a) 18
(b) 20
(c) 22
(d) 30
Answer: (a)
Explanation: Since, every student gets one chocolate. So, number of chocolates Rohit has is equal to the Number of students in the class i.e., 54. Let number of milk chocolates Rohit has = x
\( P(\text{milk chocolates}) = \frac{1}{3} \Rightarrow \frac{x}{54} = \frac{1}{3} \Rightarrow x = \frac{54}{3} = 18 \)

Question. If the probability of distributing dark chocolates is \( \frac{4}{9} \), then the number of dark chocolates Rohit has, is
(a) 18
(b) 25
(c) 24
(d) 36
Answer: (c)
Explanation: Let number of dark chocolates Rohit has = y
\( P(\text{dark chocolates}) = \frac{4}{9} \Rightarrow \frac{y}{54} = \frac{4}{9} \Rightarrow y = 24 \)

Question. The probability of distributing white chocolates is
(a) \( \frac{11}{27} \)
(b) \( \frac{8}{21} \)
(c) \( \frac{1}{9} \)
(d) \( \frac{2}{9} \)
Answer: (d)
Explanation: No. of white chocolates = \( 54 – (18 + 24) = 12 \)
\( \therefore \) Required probability = \( \frac{12}{54} = \frac{2}{9} \)

Question. The probability of distributing both milk and white chocolates is
(a) \( \frac{3}{17} \)
(b) \( \frac{5}{9} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{27} \)
Answer: (b)
Explanation: Total no. of milk & white choc. = 18 + 12 = 30
\( \therefore \) Required probability = \( \frac{30}{54} = \frac{5}{9} \)

Question. The probability of distributing all the chocolates is
(a) 0
(b) 1
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Answer: (b)
Explanation: Since every students get one chocolate each. \( \therefore \) Required probability = \( \frac{54}{54} = 1 \)

II. In a party, some children decided to play a game of musical chairs. In the game the person playing the music has been advised to stop the music at any time in the interval of 3 minutes after he starts the music in each turn. On the basis of the given information, answer the following questions.

Question. What is the probability that the music will stop within first 30 seconds after starting?
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{3} \)
Answer: (a)
Explanation: Total time = 3 mins = 3 × 60 secs = 180 secs
\( \therefore \) Required probability = \( \frac{30}{180} = \frac{1}{6} \)

Question. The probability that the music will stop within 45 seconds after starting is
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{1}{8} \)
Answer: (a)
Explanation: Required probability = \( \frac{45}{180} = \frac{1}{4} \)

Question. The probability that the music will stop after 2 minutes after starting is
(a) \( \frac{1}{8} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{3} \)
Answer: (d)
Explanation: Required probability = \( \frac{120}{180} = \frac{2}{3} \)
\( \therefore P(\text{music will stop after 2 minutes}) = 1 - \frac{2}{3} = \frac{1}{3} \)

Question. The probability that the music will not stop within first 60 seconds after starting is
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{4}{5} \)
(d) \( \frac{8}{9} \)
Answer: (b)
Explanation: Required Probability = \( 1 - \frac{60}{180} = 1 - \frac{1}{3} = \frac{2}{3} \)

Question. The probability that the music will stop within first 82 seconds after starting is
(a) \( \frac{11}{30} \)
(b) \( \frac{41}{90} \)
(c) \( \frac{31}{45} \)
(d) \( \frac{82}{90} \)
Answer: (b)
Explanation: Required probability = \( \frac{82}{180} = \frac{41}{90} \)

III. Three persons toss 3 coins simultaneously and note the outcomes. ‘Then, they ask few questions to one another. Help them in finding the answer of the following questions:

Question. The probability of getting at most one tail is
(a) 0
(b) 1
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Answer: (c)
Explanation: Total number of events = {HHH, HHT, HTT, TTH, THH, HTH, THT, TTT} = 8
Let A be the event of getting atmost one tail,
\( \therefore A = \{HHH, HHT, HTH, THH\} \)
\( \Rightarrow n(A) = 4 \therefore \) Required probability = \( \frac{4}{8} = \frac{1}{2} \)

Question. The probability of getting exactly 1 head is
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{8} \)
(d) \( \frac{3}{8} \)
Answer: (d)
Explanation: Let B be the event of getting exactly 1 head.
\( \therefore B = \{HTT, THT, TTH\} \)
\( \Rightarrow n(B) = 3 \therefore \) Required probability = \( \frac{3}{8} \)

Question. The probability of getting exactly 3 tails is
(a) 0
(b) 1
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{8} \)
Answer: (d)
Explanation: C = exactly 3 tails = {TTT} \( \Rightarrow n(C) = 1 \)
\( \therefore \) Required Probability = \( \frac{1}{8} \)

Question. The probability of getting atmost 3 heads is
(a) 0
(b) 1
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{8} \)
Answer: (b)
Explanation: D = (atmost 3 heads) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} i.e., 8
\( \therefore \) Required Probability = \( \frac{8}{8} = 1 \)

Question. The probability of getting atleast two heads is
(a) 0
(b) 1
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Answer: (c)
Explanation: E = {HHT, HTH, THH, HHH} i.e., 4
\( \therefore \) Required probability = \( \frac{n(E)}{n(S)} = \frac{4}{8} = \frac{1}{2} \)