Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
MAH CET MCA Previous Year Questions (PYQs)
MAH CET MCA Probability Distribution PYQ
MAH CET MCA PYQ
Go to Discussion
MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ
The probability distribution function of a random variable $X$ is given by
$f(x) = \dfrac{x}{18}, ; 0 \le x \le 6$
$= 0, \text{ otherwise}$
Then the value of $P(X > 2)$ is
Go to Discussion
MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ
Solution
$P(X > 2) = \int_{2}^{6} \dfrac{x}{18} , dx = \dfrac{1}{18} \left[\dfrac{x^2}{2}\right]_2^6 = \dfrac{1}{36}(36 - 4) = \dfrac{32}{36} = \dfrac{8}{9}$
MAH CET MCA PYQ
Go to Discussion
MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ
If, for a binomial distribution, the number of trials is $9$, the variance is $2$ and the probability of success is greater than that of failure, find the probability of both success.
Go to Discussion
MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ
Solution
For a binomial distribution, variance $= npq = 2$ and $n = 9$. So, $9p(1 - p) = 2 \Rightarrow 9p - 9p^2 = 2 \Rightarrow 9p^2 - 9p + 2 = 0$. $\Rightarrow p = \dfrac{9 \pm \sqrt{81 - 72}}{18} = \dfrac{9 \pm 3}{18}$. Hence, $p = \dfrac{2}{3}$ or $\dfrac{1}{3}$. Since probability of success is greater than failure, $p = \dfrac{2}{3}$.MAH CET MCA
Online Test Series,
Information About Examination,
Syllabus, Notification
and More.
View More
MAH CET MCA
Online Test Series,
Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
