BITSAT Statistics and Probability — practice questions
44 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice BITSAT Statistics and Probability in the app →Mean of n numbers x₁, x₂, ..., x_n:Median of sorted data with odd count n:Mode of data set is:Probability of certain event:Total probability of all mutually exclusive outcomes:Standard deviation of n numbers with mean x̄:Variance:For independent events A and B:Mutually exclusive events A and B:Conditional probability:Bayes' theorem:Expected value E(X) of discrete random variable:Bernoulli trial has:Binomial distribution: P(X = k) =Mean of binomial distribution with n trials, success probability p:Variance of binomial:Mean of 10 numbers is 8. If a new value 30 is added, new mean:Box has 5 red, 7 blue balls. Draw 2 without replacement. P(both red):P(both red OR both blue) in 2 draws without replacement from 5R 7B box:A coin tossed 10 times. P(exactly 5 heads):If P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2: A and B independent?For random variable X uniform on [0, 1]: E(X²) =For uniform U[0,1]: variance =E[X² + Y²] when X, Y are independent identically distributed with E(X) = 0, Var(X) = σ²:Cov(X, X) =Correlation coefficient r:5 cards from deck of 52. P(all spades):Coefficient of variation (CV):Mean of frequency distribution Σf_i x_i / Σf_i. Find mean of: 2(×3), 4(×5), 6(×2): values 2,4,6 with frequenciStandard deviation property: σ(aX + b) =For Poisson distribution with rate λ: P(X = k) =Three-card hand probability of a flush (any same suit):At least one '6' in 4 throws of a fair die:Variance of fair die roll (1-6):P(A) = 1/3, P(B) = 1/2, P(A ∩ B) = 1/6. P(A ∪ B) =From 7 men and 5 women, choose committee of 4 with at least 1 woman:A box has 6 balls, 4 are red. Two balls drawn (without replacement). P(both red):Mean Square Error (MSE) of estimator decomposes as:Properties of normal distribution: ___ % of data within 1 SD of mean.X ~ Binomial(50, 0.4). Find E[X] and Var[X]:For two events A and B, P(A | B) = 0.6, P(B) = 0.5, P(A) = 0.4. P(B | A) =Mean Absolute Deviation (MAD):Chebyshev's inequality: P(|X - μ| ≥ kσ) ≤Two independent X, Y with Var(X) = 4, Var(Y) = 9. Var(X + Y) =