Q.The mean of Binomial distribution is 20 and Standard deviation is 4. Find the parameters of the distribution.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.In a box containing 100 bulbs, 10 are defective. Write the probability that out of a sample of 5 bulbs, none is defective.

   (a) $10^{-1}$

   (b) $\left(\frac{1}{2}\right)^5$

   (c) $\left(\frac{9}{10}\right)^5$

   (d) $\frac{9}{10}$

   ›Reveal solutionSolution

   Treating each draw as an independent trial with probability $\frac{9}{10}$ of being non-defective, the probability that all $5$ are non-defective is $\left(\frac{9}{10}\right)^5$.

   Out of $100$ bulbs, $10$ are defective, so the probability a randomly chosen bulb is not defective is

   $P(\text{good})=\frac{100-10}{100}=\frac{90}{100}=\frac{9}{10}$

   …

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2. CBSE 2026Set MARCH1 markMCQ
   Q.For which value of $x$, the value of $p(x)$ of binomial distribution with parameters $n = 4$ and $p = \dfrac{1}{2}$ becomes maximum?

   (a) $0$

   (b) $2$

   (c) $3$

   (d) $4$

   ›Reveal solutionSolution

   With $n=4,\ p=\tfrac12$, $p(x)$ is maximum at $x=2$ (the mean).

   For $B(n,p)$ with $n=4,\ p=\tfrac12$, the probabilities are

   $p(x)=\binom{4}{x}\left(\tfrac12\right)^4,\quad \binom{4}{x}=1,4,6,4,1\ \text{for}\ x=0,1,2,3,4.$ …

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3. CBSE 2026Set MARCH1 mark
   Q.The probability of failure in a binomial distribution is 0.6 and the number of trials in it is 5. Find the probability of success.

   ›Reveal solutionSolution

   $p=1-q=1-0.6=0.4$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.A coin is tossed three times, then the probability to get Head at least two times will be -

   (a) $\dfrac{1}{8}$

   (b) $\dfrac{3}{8}$

   (c) $\dfrac{1}{2}$

   (d) $\dfrac{5}{8}$

   ›Reveal solutionSolution

   Use the binomial distribution with $n=3$, $p=\dfrac{1}{2}$, and add $P(X=2)+P(X=3)$.

   …

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5. CBSE 2025Set MARCH1 markMCQ
   Q.If the parameters of a binomial distribution $B(n,p)$ mean $=4$ and variance $=\dfrac{4}{3}$, the probability, $P(X\ge 5)$ is equal to :

   (a) $\left(\dfrac{1}{3}\right)^{6}$

   (b) $\left(\dfrac{2}{3}\right)^{6}$

   (c) $4\left(\dfrac{2}{3}\right)^{6}$

   (d) $\left(\dfrac{2}{3}\right)^{5}\left(\dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   Solve $np=4,\ npq=\tfrac43$ for $n=6,\ p=\tfrac23,\ q=\tfrac13$, then add $P(5)$ and $P(6)$ to get $4\left(\tfrac23\right)^6$; option (c).

   Find the parameters.

   $q=\frac{\text{variance}}{\text{mean}}=\frac{4/3}{4}=\frac13,\quad p=1-q=\frac23,\quad n=\frac{4}{p}=\frac{4}{2/3}=6.$

   Compute $P(X\ge5)=P(5)+P(6)$ with $B(6,\tfrac23)$:

   $P(5)=\binom{6}{5}\left(\frac23\right)^5\left(\frac13\right)=6\cdot\left(\frac23\right)^5\cdot\frac13=2\left(\frac23\right)^5,$

   $P(6)=\binom{6}{6}\left(\frac23\right)^6=\left(\frac23\right)^6.$

   …

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6. CBSE 2025Set MARCH1 markMCQ
   Q.In a binomial distribution, the probability of success is twice as that of failure, then out of 4 trials, the probability of no success is :

   (a) $\dfrac{2}{27}$

   (b) $\dfrac{16}{81}$

   (c) $\dfrac{1}{81}$

   (d) $\dfrac{1}{16}$

   ›Reveal solutionSolution

   From $p=2q$ and $p+q=1$ we get $q=\tfrac13$; the probability of no success in $4$ trials is $q^4=\tfrac{1}{81}$, option (c).

   Find $p$ and $q$. Given $p=2q$ and $p+q=1$:

   $2q+q=1\implies q=\frac13,\quad p=\frac23.$

   No success in $4$ trials ($X=0$): …

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7. CBSE 2025Set MARCH1 markMCQ
   Q.The binomial distribution has mean 6 and variance $\frac{12}{7}$. What will be the type of this distribution?

   (a) Positively skewed

   (b) Negatively skewed

   (c) Symmetric

   (d) Nothing can be said about the distribution

   ›Reveal solutionSolution

   Since $q = \dfrac{\text{variance}}{\text{mean}} = \dfrac{2}{7}$ gives $p = \dfrac{5}{7} > 0.5$, the binomial distribution is negatively skewed — option (b).

   GSEB Class-12 Statistics, Binomial Distribution:

   For a binomial distribution, mean $= np$ and variance $= npq$. Therefore:

   $q = \frac{\text{variance}}{\text{mean}} = \frac{12/7}{6} = \frac{12}{42} = \frac{2}{7}$

   $p = 1 - q = 1 - \frac{2}{7} = \frac{5}{7} \approx 0.714$

   …

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8. CBSE 2025Set MARCH1 mark
   Q.Mean of a symmetrical binomial distribution is 9. Find the value of its parameter $n$.

   ›Reveal solutionSolution

   For a symmetric binomial distribution $p = q = \tfrac12$; then $np = 9 \Rightarrow n = 18$.

   GSEB Class-12 Statistics, Binomial Distribution:

   A binomial distribution is symmetric only when $p = q = \dfrac{1}{2}$.

   …

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9. CBSE 2025Set 465/S/WXYZ/41 markMCQ
   Q.If the mean and standard deviation of a binomial distribution are 12 and 2 respectively, then the value of the parameter p is : (A) $\dfrac{5}{6}$ (B) $\dfrac{1}{6}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   From mean $np=12$ and variance $npq=4$, we get $q=\tfrac13$, so $p=1-q=\tfrac23$.

   Binomial: mean $=np$, variance $=npq$, standard deviation $=\sqrt{npq}$, with $q=1-p$.

   1. Mean: $np = 12$.
   2. Standard deviation $=2 \Rightarrow$ variance $npq = 2^2 = 4$. …

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10. CBSE 2024Set D1 markMCQ
    Q.A coin is tossed 10 times. The probability of getting exactly six heads is

    (a) $^{10}C_6\left(\frac{1}{2}\right)^6$

    (b) $^{10}C_6\left(\frac{1}{2}\right)^7$

    (c) $^{10}C_6\left(\frac{1}{2}\right)^8$

    (d) $^{10}C_6\left(\frac{1}{2}\right)^{10}$

    ›Reveal solutionSolution

    Binomial: $P(X=6)=\,^{10}C_6(\tfrac12)^6(\tfrac12)^4=\,^{10}C_6(\tfrac12)^{10}$.

    A coin toss is a Bernoulli trial with $p=P(\text{head})=\tfrac12$. For $n=10$ tosses, the number of heads follows a binomial distribution:

    …

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11. CBSE 2024Set ANNUAL1 mark
    Q.If in a binomial distribution mean is 5 and variance is 4, then write the number of trials.

    ›Reveal solutionSolution

    For a binomial distribution, mean $= np$ and variance $= npq$. Dividing gives $q$, then $p$, then $n$.

    For a binomial distribution $B(n,p)$, mean $= np = 5$ and variance $= npq = 4$.

    Dividing variance by mean: $q = \dfrac{npq}{np} = \dfrac{4}{5}$.

    …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The mean and variance of a binomial distribution are 3 and 3/2, then P(X≥1) is -

    (a) 64/63

    (b) 63/64

    (c) 1/64

    (d) 1/63

    ›Reveal solutionSolution

    Use mean $=np$, variance $=npq$ to find $n$ and $p$, then compute $P(X\ge1) = 1-P(X=0)$.

    Given $np = 3$ and $npq = \frac{3}{2}$.

    Dividing: $q = \frac{npq}{np} = \frac{3/2}{3} = \frac{1}{2}$, so $p = 1-q = \frac{1}{2}$.

    From $np=3$: $n\cdot\frac{1}{2} = 3 \Rightarrow n=6$

    …

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