Ratio Mathematica Volume 40, 2021, pp. 123-137 123 Reliability estimation of Weibull- exponential distribution via Bayesian approach Arun Kumar Rao* Himanshu Pandey† Abstract Weibull-exponential distribution is considered. Bayesian method of estimation is employed in order to estimate the reliability function of Weibull-exponential distribution by using non-informative and beta priors. In this paper, the Bayes estimators of the reliability function have been obtained under squared error, precautionary and entropy loss functions. Keywords: Weibull-exponential distribution. Reliability. Bayesian method. Non-informative and beta priors. Squared error, precautionary and entropy loss functions. 2010 AMS subject classification: 60E05, 62E15, 62H10, 62H12.§ * Department of Statistics, Maharana Pratap P.G. College, Jungle Dhusan, Gorakhpur, INDIA; arunrao1972@gmail.com. † Department of Mathematics & Statistics, DDU Gorakhpur University, Gorakhpur, INDIA; himanshu_pandey62@yahoo.com. §Received on January 12th, 2021. Accepted on May 12th, 2021. Published on June 30th, 2021. doi: 10.23755/rm.v40i1.570. ISSN: 1592-7415. eISSN: 2282-8214. ©Rao and Pandey. This paper is published under the CC-BY licence agreement. A.K.Rao and H.Pandey 136 1. Introduction The Weibull-exponential distribution was proposed by Oguntunde et al. [1]. They obtained some of its basic Mathematical properties. This distribution is useful as a life testing model and is more flexible than the exponential distribution. The probability function ( )f x; and distribution function ( )F x; of Weibull- exponential distribution are respectively given by ( ) ( ) ( ) 1 1 1 0 a a x a x x f x; a e e exp e ; x .       − −  = − − −     (1) ( ) ( )1 1 0 0 a x e F x; e ; x , .     − − − = −   (2) Let ( )R t denote the reliability function, that is, the probability that a system will survive a specified time t comes out to be ( ) ( )1 0 0 a t e R t e ; t , .    − − =   (3) And the instantaneous failure rate or hazard rate, h(t) is given by ( ) ( )1a t th t a e e .  −= − (4) From equation (1) and (3), we get ( )( ) ( ) ( ) ( ) ( ) ( ) 1 1 11 1 1 a x t a x e a x e a t a e f x; R t e log R t R t ; 0