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PUCET (Punjab) Binomial Theorem — practice questions

Explore 136 free Binomial Theorem questions in PUCET (Punjab) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Binomial Theorem questions and exercises with answers

Showing questions 1–30 of 136.

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 →$\sin(180° - \theta)$ equals:Read the answer and worked solution →$\sin^2\theta + \cos^2\theta$ equals:Read the answer and worked solution →$\sin(A + B) = $Read the answer and worked solution →$\cos 2\theta$ equals:Read the answer and worked solution →If $\sin\theta + \cos\theta = 1$, then $\sin\theta \cdot \cos\theta$ equals:Read the answer and worked solution →$\cos 3\theta$ equals:Read the answer and worked solution →A **random variable** is best defined as:Read the answer and worked solution →A **discrete** random variable is one whose possible values:Read the answer and worked solution →For a continuous random variable $X$, the probability that $X$ equals any specific value is:Read the answer and worked solution →The **cumulative distribution function (c.d.f.)** $F(x)$ of a discrete random variable is:Read the answer and worked solution →When two fair dice are rolled and $X$ denotes the sum of the upper faces, the range of $X$ is:Read the answer and worked solution →If a fair coin is tossed twice and $X$ = number of heads, then $P[X = 1]$ equals:Read the answer and worked solution →If two cards are drawn with replacement from a deck of 52, $P[X = 2]$ where $X$ = number of aces equals:Read the answer and worked solution →A fair die is thrown. Let $X$ = number of factors of the number on the upper face. $P[X = 2]$ equals:Read the answer and worked solution →$f(x) = 3x^2$ for $0 < x < 1$ is the p.d.f. of $X$. The probability $P[1/2 < X < 1]$ is:Read the answer and worked solution →If $f(x) = kx^2(1 - x)$ for $0 < x < 1$ is the p.d.f. of $X$, the value of $k$ is:Read the answer and worked solution →For p.d.f. $f(x) = x/8$, $0 < x < 4$, the probability $P[x < 1.5]$ is:Read the answer and worked solution →Find $k$ if $f(x) = kx$ for $0 < x < 2$ and 0 otherwise is a valid p.d.f.Read the answer and worked solution →The c.d.f. $F(x)$ of the p.d.f. $f(x) = 3x^2$ for $0 < x < 1$ is:Read the answer and worked solution →Two cards are drawn with replacement. $X$ = number of aces. The probability $P[X \geq 1]$ equals:Read the answer and worked solution →For p.d.f. $f(x) = x^2/3$ for $-1 < x < 2$, the probability $P[0 < X \leq 1]$ equals:Read the answer and worked solution →For p.d.f. $f(x) = k(4 - x^2)$ for $-2 \leq x \leq 2$ (and 0 elsewhere), the value of $k$ is:Read the answer and worked solution →For the c.d.f. $F(x) = 3x - 2x^3$ (from p.d.f. $f(x) = 3(1 - 2x^2)$ on $[0, 1]$), $F(0.5)$ equals:Read the answer and worked solution →If a fair coin is tossed $n$ times, the probability of any exact pattern (like all heads) is:Read the answer and worked solution →If a die is thrown and 'getting an odd number' is success, then $p$ and $q$ are:Read the answer and worked solution →A bag has balls marked 0–9. Four balls drawn with replacement. Probability that **none** is marked '0' is:Read the answer and worked solution →Probability of bomb hitting target = 0.8. Out of 10 bombs dropped, probability that **exactly 2 miss** equals (with $p = 0.2$ miss):Read the answer and worked solution →sin(0°) equals: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. $8$
Show answer and solution

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. $\dfrac{2}{3}$
Show answer and solution

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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$\sin(180° - \theta)$ equals:

  1. $-\cos\theta$
  2. $-\sin\theta$
  3. $\sin\theta$
  4. $\cos\theta$
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Correct answer: C. $\sin\theta$

Supplementary-angle identity: $\sin(180° - \theta) = \sin\theta$. Reason: sine is positive in the second quadrant (where $180° - \theta$ lies for $0° < \theta < 180°$), and the reference angle is $\theta$. Companion identities: - $\cos(180° - \theta) = -\cos\theta$ - $\tan(180° - \theta) = -\tan\theta$

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