Topics

General Knowledge
General Science
General English
Aptitude
General Computer Science
General Intellingence and Reasoning
Current Affairs
Exams
Elementary Mathematics
English Literature

Which lone judge among the five-judge constitution bench gave a dissenting opinion in the SC's 4: 1 majority verdict allowing women's entry in Kerala's Sabarimala temple?

) Indu Malhotra
) DYChandrachud
) Rohinton Nariman
) Ashok Bhushan

Correct Answer:

) Indu Malhotra

Description for Correct answer:
A five-judge Constitution Bench of the Supreme Court, in a majority judgment, on Friday upheld the right of women of all ages to worship in places of their choice. This verdict will pave the way for the famed temple of Lord Ayyappa in Sabarimala, Kerala, to open its doors to women in the menstruation age also.

Part of solved Current Affairs-October-2018 questions and answers : >> Current Affairs-October-2018

Comments

Similar Questions

1). Which state topped the 'Ease of Living Index' rankings launched by the Ministry of Housing and Urban Affairs?

A). Kerala

B). Telangana

C). Andhra Pradesh

D). Rajasthan

2). Which movie has been chosen as the India's official entry to Oscars 2019? (A)

A). Village Rockstars

B). October

C). Kadvi Hawa

D). Raazi

3). If u and v are independent Chi-square variables with \(r_{1}\) and \(r_{2}\) degrees of freedom, then the distribution of \(w=\frac{u/r_{1}}{v/r_{2}}\) is called

A). Chi-square distribution

B). t-distribution

C). F-distribution

D). Poisson distribution

4). The p.d.f. of a Chi-square distribution is

A). \( \Large \frac{1}{\Gamma \left(^{r}/_{2}\right)2^{^{r}/_{2}}}x^{^{r}/_{2^{-1}}}e^{^{-x}/_{2}} \)

B). \( \Large \frac{1}{\Gamma \left(^{r}/_{2}\right)2^{^{r}/_{2}}}x^{^{r}/_{2^{-1}}} \)

C). \( \Large \frac{1}{\Gamma \left(^{r}/_{2}\right)2^{^{r}/_{2}}}e^{^{-x}/_{2}} \)

D). \( \Large \frac{1}{\Gamma \left(^{r}/_{2}\right)2^{^{r}/_{2}}}x^{^{r}/_{2}}e^{^{-x}/_{2}} \)

5). Finding the relation between two random variables is called

A). regression

B). analysis

C). sample tests

D). binomial

6). Let \(X_{ij}, i=1,2 \) and \(j = 1,2,3\) denote \(n=6\) random variables that are independent and gave normal distribution with common variance \(\sigma^{2}\). The means of these normal distributions are \(\mu_{ij}=\mu+ \alpha _{i}+ \beta _{j}\) where \(\sum\limits_{i=1}^{2} \alpha _{i}=0,\sum\limits_{j=1}^{3} \beta _{j}=0\) This concept is associated with

A). Sampling

B). Correlation

C). Regression

D). Analysis of variance

7). The arithmetic mean of Poisson distribution is

A). \(\frac{\lambda^{r}e^{-\lambda}}{\underline{|r}}\)

B). \(e^{-\lambda}\)

C). \(\lambda\)

D). \(\lambda^{r}\)

8). A lot consists of 100 fuses. Five of these fuses are chosen at random and tested; if all 5 "blow" at the correct amperage, the lot is accepted. In fact, there are 20 defective fuses in the lot. Let the random variable X be the number of defective fuses among the 5 that are inspected. The random variable X is an example of

A). Beta distribution

B). Hypergeometric distribution

C). Poisson distribution

D). Gamma distribution

9). Let \(X_{1}, X_{2}\) be a random sample of size \(n=2\) from a .Standard normal distribution. Let \(Y_{1}=X_{1}/X_{2}\) and \(Y_{2}X_{2}\). Now the marginal p.d.f. of \(Y_{1}\) is that of a

A). Poisson distribution

B). Negative binomial distribution

C). Multinomial distribution

D). Cauchy distribution

10). Let the random variable X have a distribution with finite variance \(\sigma^{2}\) and mean \(\mu\). Then for every \(k > 0\), the Chebyshev's inequality is

A). \(P_{r} \left(|X-\mu| \ge k\sigma\right)\le \frac{1}{k^{2}} \)

B). \(P_{r} \left(|X-\mu| \ge k\sigma\right)\ge \frac{1}{k^{2}} \)

C). \(P_{r} \left(|X-\mu| \ge k\sigma\right)\le 1-\frac{1}{k^{2}} \)

D). \(P_{r} \left(|X-\mu| < k\sigma\right)\le 1-\frac{1}{k^{2}} \)