KC+CET PYQs for Binomial Distribution with Solutions: Practice KCET Previous Year Questions

Updated on - Feb 16, 2026

Binomial Distribution is an important topic in the Mathematics section in KCET exam. Practising this topic will increase your score overall and make your conceptual grip on KCET exam stronger.

This article gives you a full set of KCET PYQs for Binomial Distribution with explanations for effective preparation. Practice of KCET Mathematics PYQs including Binomial Distribution questions regularly will improve accuracy, speed, and confidence in the KCET 2026 exam.

KCET PYQs for Binomial Distribution with Solutions

1.
If a random variable X follows the binomial distribution with parameters n = 5, p and P (X=2) =9P (X=3), then p is equal to
- 10
- $\frac{1}{10}$
- 5
- $\frac{1}{5}$
View Solution
2.
If a random variable $X$ follows the binomial distribution with parameters $n = 5$, $p$, and $P(X = 2) = 9P(X = 3)$, then $p$ is equal to:
- $10$
- $\frac{1}{10}$
- $5$
- $1\frac{1}{5}$
View Solution
3.
The probability of solving a problem by three persons $A, B$ and $C$ independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is
- $\frac{1}{12}$
- $\frac{1}{4}$
- $\frac{1}{24}$
- $\frac{1}{8}$
View Solution
4.
The digit in the unit's place of $7^{171} +(177)!$ is
- 3
- 2
- 1
- 0
View Solution

View more KCET Questions

SHARE ON FACEBOOK SHARE ON TWITTER

Comments

No Comments To Show

SUBSCRIBE TO OUR NEWS LETTER

COLLEGE NOTIFICATIONS

EXAM NOTIFICATIONS

NEWS UPDATES

Download the Collegedunia app on

KCET PYQs for Binomial Distribution with Solutions

1.If a random variable X follows the binomial distribution with parameters n = 5, p and P (X=2) =9P (X=3), then p is equal to

2.If a random variable $X$ follows the binomial distribution with parameters $n = 5$, $p$, and $P(X = 2) = 9P(X = 3)$, then $p$ is equal to:

3.The probability of solving a problem by three persons $A, B$ and $C$ independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is

4.The digit in the unit's place of $7^{171} +(177)!$ is

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If a random variable X follows the binomial distribution with parameters n = 5, p and P (X=2) =9P (X=3), then p is equal to

2.
If a random variable $X$ follows the binomial distribution with parameters $n = 5$, $p$, and $P(X = 2) = 9P(X = 3)$, then $p$ is equal to:

3.
The probability of solving a problem by three persons $A, B$ and $C$ independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is

4.
The digit in the unit's place of $7^{171} +(177)!$ is