0 votes 0 votes If $A_1, A_2, \dots , A_n$ are independent events with probabilities $p_1, p_2, \dots , p_n$ respectively, then $P( \cup_{i=1}^n A_i)$ equals $\Sigma_{i=1}^n \: \: p_i$ $\Pi_{i=1}^n \: \: p_i$ $\Pi_{i=1}^n \: \: (1-p_i)$ $1-\Pi_{i=1}^n \: \: (1-p_i)$ Probability isi2016-mmamma probability independent-events + – go_editor asked Sep 13, 2018 recategorized Nov 19, 2019 by Lakshman Bhaiya go_editor 298 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Consider value of n as 2, We know P(A1 U A2) = P(A1) + P(A2) - P(A1nA2) =p1 + p2 - p1p2 (Ai are Independent Events) =p1(1-p2) + p2 =p1(1-p2)+p2+1-1 (Adding and Subtracting 1) =p1(1-p2)-(1-p2)+1 = 1+(1-p2)(p1-1) = 1-(1-p1)(1-p2) Answer : D SatyamK answered Mar 19, 2020 SatyamK comment Share Follow See all 0 reply Please log in or register to add a comment.