Class 10 Probability Mock Test (CBSE)

Topics in Probability

• Theoretical probability
• Complementary events
• Impossible and sure events

Probability practice questions with answers

1. Q1Multiple choice·1 mark
   A box contains $3$ blue, $2$ white, and $4$ red marbles. If three marbles are drawn at random without replacement, the probability of getting all different colored marbles is:

   1. (a)$\frac{2}{7}$
   2. (b)$\frac{1}{7}$
   3. (c)$\frac{4}{21}$
   4. (d)$\frac{12}{35}$
   Show answer and step marks

   1. Correct option: (a) $\frac{2}{7}$. Total ways to draw $3$ marbles out of $9$ is $^9C_3 = 84$, and ways to get one of each color is $^3C_1 \times ^2C_1 \times ^4C_1 = 24$, giving probability $\frac{24}{84} = \frac{2}{7}$.+1

   Answer: (a)
2. Q2Assertion and reason·1 mark
   Assertion (A): When a fair die is rolled, the probability of the complementary event of getting an even number is $\frac{1}{3}$. Reason (R): For any event $E$, $P(E) + P(\overline{E}) = 1$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (d) Assertion (A) is false but Reason (R) is true.+1

   Answer: (d)
3. Q3Assertion and reason·1 mark
   Assertion (A): If the probability of an event happening is $0.35$, then the probability of its complementary event is $0.65$. Reason (R): For any event $E$, $P(E) + P(\text{not } E) = 1$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Both Assertion and Reason are true, and the Reason correctly explains the formula used to find the complementary probability ($1 - 0.35 = 0.65$).+1

   Answer: (a)
4. Q4Multiple choice·1 mark
   A bag contains $150$ cards numbered $1$ to $150$. A card is drawn at random from the bag. What is the probability that the selected card bears a number which is a perfect square or a multiple of $7$, but not both?

   1. (a)$\frac{31}{150}$
   2. (b)$\frac{16}{75}$
   3. (c)$\frac{11}{50}$
   4. (d)$\frac{17}{75}$
   Show answer and step marks

   1. Correct option: (a) $\frac{31}{150}$. Total outcomes = $150$. Perfect squares = $12$, multiples of $7$ = $21$, and their intersection is $1$ ($49$). Favorable outcomes for numbers that are either a perfect square or a multiple of $7$ but not both = $(12 + 21) - 2(1) = 31$, giving probability $\frac{31}{150}$.+1

   Answer: (a)
5. Q5Short answer·2 marks
   Two identical fair coins are tossed together once. Find the probability of getting exactly one head.
   Show answer and step marks

   1. The equally likely outcomes are HH, HT, TH and TT, so the total number of outcomes is 4.+1
   2. Outcomes with exactly one head are HT and TH, so 2 favourable outcomes and P(exactly one head) = $\frac{2}{4} = \frac{1}{2}$.+1

   Answer: $\frac{1}{2}$
6. Q6Long answer·5 marks
   Two fair dice, one red and one blue, are thrown together once. Let E be the event that the sum of the numbers on the two dice is at least 10. (a) List all the outcomes in E as ordered pairs (red die, blue die). [1] (b) Find P(E). [1] (c) Write, in words, the complementary event of E, denoted $\bar{E}$. [1] (d) Find P($\bar{E}$). [1] (e) Give a reason why P(E) + P($\bar{E}$) = 1. [1]
   Show answer and step marks

   1. (a) Outcomes in E: (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)+1
   2. (b) Total outcomes = 36, so P(E) = 6/36 = 1/6+1
   3. (c) $\bar{E}$: the sum of the numbers on the two dice is less than 10+1
   4. (d) P($\bar{E}$) = 1 − P(E) = 1 − 1/6 = 5/6+1
   5. (e) E and $\bar{E}$ cannot occur together and one of them must occur, so their probabilities add to 1+1

   Answer: (a) (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6); (b) P(E) = 1/6; (c) the sum is less than 10; (d) P($\bar{E}$) = 5/6; (e) E and $\bar{E}$ are mutually exclusive and exhaustive, so P(E) + P($\bar{E}$) = 1/6 + 5/6 = 1.

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