Q.A fair coin is tossed 5 times. Find the probability of getting exactly 3 heads.

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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