The probability that a bomb dropped from a plane strikes the target is $\dfrac{4}{5}$. What is the probability that out of 6 bombs dropped, exactly 2 bombs strike the target ? (A) $2\left(\dfrac{4}{5}\right)^5$ (B) $1 - 2\left(\dfrac{4}{5}\right)^5$ (C) $\dfrac{48}{5^5}$ (D) $\dfrac{64}{5^6}$

> [!TLDR] > $P(X=2)=\binom{6}{2}(4/5)^2(1/5)^4 = \dfrac{48}{5^5}$. > [!FORMULA] > Binomial: $P(X=r)=\binom{n}{r}p^{r}q^{n-r}$, here $n=6$, $p=\tfrac45$ (strike), $q=\tfrac15$ (miss), $r=2$. 1. Number of ways: $\binom{6}{2}=15$. 2. Probability: $P(X=2)=15\left(\dfrac45\right)^2\left(\dfrac15\right)^4 = 15\cdot\dfrac{16}{25}\cdot\dfrac{1}{625}$. …

The probability that a bomb dropped from a plane strikes the target is (4)/(5). What is the probability that o

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\dfrac{4}{5}

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Concept understanding — Binomial Distribution

Binomial Distribution: From Intuition to Precision

Imagine you flip a fair coin 10 times. You want to know the probability of getting exactly 6 heads. This is the kind of question the binomial distribution answers — it models the number of "successes" in a fixed number of independent trials, where each trial has only two outcomes.

The Intuition

Think of a single trial. You roll a die and call "getting a 4" a success. That's one trial. Now roll the die 5 times. The number of times you get a 4 could be 0, 1, 2, 3, 4, or 5. Each roll is independent — the result of one roll doesn't affect the next. The probability of success (getting a 4) stays the same every time: $p = \frac{1}{6}$ .

The binomial distribution tells you exactly how likely each possible count of successes is, given:

a fixed number of trials $n$ ,
a constant success probability $p$ per trial,
independent trials.

The Key Conditions

For a situation to be modelled by a binomial distribution, all four must hold:

Fixed number of trials ( $n$ ). You decide in advance how many times you'll repeat the experiment.
Two outcomes per trial — "success" and "failure". These are just labels; success is whatever outcome you're counting.
Constant probability of success ( $p$ ) on every trial.
Independent trials. The outcome of one trial does not influence another.

Watch out

A common mistake is to apply the binomial distribution when trials are not independent — for example, drawing cards without replacement from a deck. That's a hypergeometric situation, not binomial.

The Precise Statement

Let $X$ be the number of successes in $n$ independent trials, each with success probability $p$ . Then $X$ follows a binomial distribution with parameters $n$ and $p$ , written:

X \sim \text{Binomial}(n, p)

The probability of getting exactly $k$ successes (where $k = 0, 1, 2, \dots, n$ ) is:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Breaking Down the Formula

Three pieces multiply together:

$p^k$ — the probability that the $k$ successes happen. Since each success has probability $p$ , and there are $k$ of them, this factor is $p$ multiplied by itself $k$ times.
$(1-p)^{n-k}$ — the probability that the remaining $n-k$ trials are failures. Each failure has probability $1-p$ .
$\binom{n}{k}$ — the number of ways to choose which $k$ of the $n$ trials are the successes. The successes could be the first $k$ trials, or the last $k$ , or any arrangement. This binomial coefficient counts all possible arrangements.

Tip

The binomial coefficient $\binom{n}{k}$ is read as "n choose k" and equals $\frac{n!}{k!(n-k)!}$ . For example, $\binom{5}{2} = \frac{5!}{2!3!} = 10$ .

A Worked Example

Problem: A multiple-choice test has 10 questions, each with 4 options. You guess every answer. What is the probability you get exactly 3 correct?

Solution:

$n = 10$ (10 trials)
$p = \frac{1}{4}$ (probability of guessing correctly on one question)
$k = 3$ (we want exactly 3 successes)

P(X = 3) = \binom{10}{3} \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^{7}

Compute $\binom{10}{3} = 120$ , so:

P(X = 3) = 120 \times \frac{1}{64} \times \left(\frac{3}{4}\right)^7

$\left(\frac{3}{4}\right)^7 = \frac{2187}{16384}$ , so:

P(X = 3) = 120 \times \frac{1}{64} \times \frac{2187}{16384} = \frac{120 \times 2187}{64 \times 16384}

Simplify: $120/64 = 15/8$ , so:

P(X = 3) = \frac{15 \times 2187}{8 \times 16384} = \frac{32805}{131072} \approx 0.2503

So there's about a 25% chance of getting exactly 3 correct by pure guessing.

Mean and Variance

For a binomial distribution, two important summary measures are: …

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$P(X=2)=\binom{6}{2}(4/5)^2(1/5)^4 = \dfrac{48}{5^5}$ .

P(X=r)=\binom{n}{r}p^{r}q^{n-r}

Number of ways: $\binom{6}{2}=15$ .
Probability: $P(X=2)=15\left(\dfrac45\right)^2\left(\dfrac15\right)^4 = 15\cdot\dfrac{16}{25}\cdot\dfrac{1}{625}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

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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CBSE 2026Set MARCH1 markMCQ
$x$
(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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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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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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CBSE 2025Set MARCH1 markMCQ
$B(n,p)$
(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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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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CBSE 2025Set MARCH1 markMCQ
$\frac{12}{7}$
(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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CBSE 2025Set MARCH1 mark
$n$
›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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CBSE 2025Set 465/S/WXYZ/41 markMCQ
$\dfrac{5}{6}$
›Reveal solutionSolution
From mean $np=12$ and variance $npq=4$ , we get $q=\tfrac13$ , so $p=1-q=\tfrac23$ .

$=np$

Mean: $np = 12$ .

Standard deviation $=2 \Rightarrow$ variance $npq = 2^2 = 4$ . …
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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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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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CBSE 2024Set ANNUAL1 markMCQ
$Q. The mean and variance of a binomial distribution are 3 and 3/2, then P(X\geq1) is -$
(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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