A bag consists of 10 balls each marked with one of the digits from 0 to 9. If 4 balls are drawn successively w

Question of 165

Q. A bag consists of 10 balls each marked with one of the digits from 0 to 9. If 4 balls are drawn successively with replacement from the bag, what is the probability that one ball is marked with the digit 1?

Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2023Subjective· 4mImportance★★★★★

0% · 0/165 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

Model as a binomial experiment with $n=4$ trials, success probability $p=1/10$ (drawing digit 1), and find $P(X=1)$ .

Since the balls are drawn with replacement, each of the $4$ draws is an independent Bernoulli trial with the same success probability. "Success" = drawing the ball marked $1$ : since there are $10$ equally likely balls, $p = \dfrac{1}{10}$ , and $q = 1-p = \dfrac{9}{10}$ .

Let $X$ = number of times the ball marked $1$ is drawn in $4$ trials. Then $X \sim \text{Binomial}(n=4, p=1/10)$ , with

$P(X=r) = \binom{4}{r}p^r q^{4-r}.$

…

Start your 14-day free trial to unlock the full solution →

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

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

…

Start your 14-day free trial to see the full solution →
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

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

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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 $P(X=2)+P(X=3)$ .

…

Start your 14-day free trial to see the full solution →
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 $B(6,\tfrac23)$ :

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

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

…

Start your 14-day free trial to see the full solution →
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 ( $X=0$ ): …

Start your 14-day free trial to see the full solution →
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:

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

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

…

Start your 14-day free trial to see the full solution →
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 $p = q = \dfrac{1}{2}$ .

…

Start your 14-day free trial to see the full solution →
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 $npq = 2^2 = 4$ . …
Start your 14-day free trial to see the full solution →
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:

…

Start your 14-day free trial to see the full solution →
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: $q = \dfrac{npq}{np} = \dfrac{4}{5}$ .

…

Start your 14-day free trial to see the full solution →
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 $np=3$ : $n\cdot\frac{1}{2} = 3 \Rightarrow n=6$

…

Start your 14-day free trial to see the full solution →

←Previous chapterCh 12Linear Programming

Q.A bag consists of 10 balls each marked with one of the digits from 0 to 9. If 4 balls are drawn successively with replacement from the bag, what is the probability that one ball is marked with the digit 1?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days