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KCET Probability — practice questions

Explore 59 free Probability questions in KCET Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Probability questions and exercises with answers

Showing questions 1–30 of 59.

A fair coin is tossed twice. How many outcomes are there in the sample space?Read the answer and worked solution →A fair six-sided die is rolled once. What is the probability of getting an even number?Read the answer and worked solution →Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.Read the answer and worked solution →The arithmetic mean of n values x1, x2, ..., xn is:Read the answer and worked solution →Median of an odd-length sorted data set is:Read the answer and worked solution →Probability of a certain event:Read the answer and worked solution →Probability of rolling a 6 on a fair die:Read the answer and worked solution →For independent events A and B, P(A and B) equals:Read the answer and worked solution →For mutually exclusive events A and B, P(A or B) equals:Read the answer and worked solution →Find mean of {2, 4, 6, 8, 10}:Read the answer and worked solution →Mode of {1, 2, 2, 3, 4, 4, 4, 5}:Read the answer and worked solution →Variance of {2, 4, 6, 8, 10}:Read the answer and worked solution →Standard deviation of {2, 4, 6, 8, 10}:Read the answer and worked solution →P(at least one head in 2 fair coin tosses):Read the answer and worked solution →Probability of drawing a heart from a standard 52-card deck:Read the answer and worked solution →P(drawing an Ace OR King from a deck) =Read the answer and worked solution →Two dice rolled. P(sum = 7) =Read the answer and worked solution →Bayes' theorem relates P(A|B) and P(B|A):Read the answer and worked solution →Number of ways to arrange 5 distinct books on a shelf:Read the answer and worked solution →Number of ways to choose 3 students from a class of 10:Read the answer and worked solution →A box has 5 red, 3 blue marbles. Probability of drawing 2 red and 1 blue in any order (without replacement, 3 draws):Read the answer and worked solution →For a Poisson distribution with mean λ = 3, P(X = 2):Read the answer and worked solution →Standard deviation has the same units as:Read the answer and worked solution →For 50 students with mean score 60 and variance 25, the coefficient of variation (sigma/mean × 100%):Read the answer and worked solution →P(picking a king OR a heart from 52 cards):Read the answer and worked solution →Two events A and B: P(A) = 0.3, P(B) = 0.4, P(A and B) = 0.1. Are they independent?Read the answer and worked solution →5 cards drawn from a standard deck. Probability all 5 are aces:Read the answer and worked solution →Number of distinguishable arrangements of letters in 'STATISTICS':Read the answer and worked solution →Conditional probability P(A | B) is defined as:Read the answer and worked solution →In a survey, 60% read newspaper A, 40% read B, 20% read both. P(reading neither):Read the answer and worked solution →

Worked practice questions

A fair coin is tossed twice. How many outcomes are there in the sample space?

  1. $2$
  2. $4$
  3. $6$
  4. $8$
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Correct answer: B. $4$

Each toss has 2 possible outcomes (H or T), and the tosses are independent. By the multiplication principle, the sample space has $2 \times 2 = 4$ outcomes: {HH, HT, TH, TT}.

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A fair six-sided die is rolled once. What is the probability of getting an even number?

  1. $\dfrac{1}{6}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{2}{3}$
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Correct answer: B. $\dfrac{1}{2}$

The even outcomes are {2, 4, 6}, so the number of favourable outcomes is 3. The total sample space has 6 outcomes. $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$.

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Events $A$ and $B$ are mutually exclusive with $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$.

  1. $0.12$
  2. $0.70$
  3. $1.00$
  4. $0.58$
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Correct answer: B. $0.70$

Mutually exclusive events cannot occur together, so $P(A \cap B) = 0$. The general addition rule simplifies: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0 = 0.7$. Option A ($0.12$) is the product $P(A) \cdot P(B)$, which is the joint probability for *independent* events, not mutually exclusive ones. **The distinction matters:** mutually exclusive means $P(A \cap B) = 0$ independent means $P(A \cap B) = P(A) \cdot P(B)$. These two assumptions cannot both hold unless one event has zero probability.

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