alleryP(drawing an Ace OR King from a deck) =
- 1/13
- 1/26
- 2/13
- 4/13
Show answer and solution
Correct answer: C. 2/13
4 aces, 4 kings, mutually exclusive: P = (4 + 4)/52 = 8/52 = 2/13.
Bihar Board Class 12 (Arts) Probability — practice questions
Explore 45 free Probability questions in Bihar Board Class 12 (Arts) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 1–30 of 45.
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 →Binomial distribution: P(X = k) for n trials, success probability p:Read the answer and worked solution →E(X) and Var(X) for fair coin tossed n times (success = heads, p = 0.5):Read the answer and worked solution →Two cards drawn from a deck (no replacement). P(both kings):Read the answer and worked solution →The probability that a marksman will hit a target is given as $\frac{1}{5}$. Then the probability that at least once hit in 10 shots isRead the answer and worked solution →The number of elements in the sample space when a coin is tossed twice is:Read the answer and worked solution →The probability of a sure event is:Read the answer and worked solution →The probability of an impossible event is:Read the answer and worked solution →For any event A, $P(\text{not }A)$ equals:Read the answer and worked solution →If A and B are mutually exclusive events, then $P(A\cap B)$ equals:Read the answer and worked solution →For any two events A and B, $P(A\cup B)$ equals:Read the answer and worked solution →In the classical definition, the probability of an event E is:Read the answer and worked solution →Three fair coins are tossed simultaneously. The probability of getting EXACTLY two heads isRead the answer and worked solution →A card is drawn from a standard deck of 52. The probability that it is a KING or a QUEEN isRead the answer and worked solution →If $P(A) = 0.4$ and $P(B) = 0.5$ and $P(A \cap B) = 0.2$, then $P(A \cup B)$ isRead the answer and worked solution →What is the probability that a randomly drawn card from a 52-card deck is a FACE CARD (King, Queen, or Jack)?Read the answer and worked solution →
Worked practice questions
Two dice rolled. P(sum = 7) =
- 1/36
- 1/6
- 6/36
- Many
Show answer and solution
Correct answer: B. 1/6
**Favorable outcomes:** (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes.
**Total:** 36. P = 6/36 = 1/6. Sum of 7 is the most likely sum on 2 dice.
Bayes' theorem relates P(A|B) and P(B|A):
- P(A|B) = P(B|A)
- P(A|B) = P(B|A) × P(A) / P(B)
- P(A|B) = P(A) + P(B)
- Cannot relate
Show answer and solution
Correct answer: B. P(A|B) = P(B|A) × P(A) / P(B)
**Bayes' theorem:** P(A|B) = P(B|A) × P(A) / P(B).
Crucial for inverse probability problems (e.g., disease testing given a positive result).
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