20 votes 20 votes Two dice are thrown simultaneously. The probability that at least one of them will have $6$ facing up is $\frac{1}{36}$ $\frac{1}{3}$ $\frac{25}{36}$ $\frac{11}{36}$ Probability gate1996 probability easy + – Kathleen asked Oct 9, 2014 • retagged Jun 26, 2017 by Silpa Kathleen 4.9k views answer comment Share Follow See 1 comment See all 1 1 comment reply Manu Thakur commented Jan 8, 2018 reply Follow Share Sample space $=6*6 = 36$ Favorable cases $= (6,6), (6,1)(6,2)(6,3)(6,4)(6,5)(1,6)(2,6)(3,6)(4,6)(5,6)$ $=\frac{11}{36}$ 15 votes 15 votes Please log in or register to add a comment.
Best answer 24 votes 24 votes Answer is (D) $1- ($no. $6$ in both the dice $)=1-\left(\dfrac{5}{6} \times\dfrac{5}{6}\right)=\dfrac{11}{36}.$ Bhagirathi answered Oct 16, 2014 • edited Sep 25, 2018 by Krithiga2101 Bhagirathi comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments sourav. commented Jan 30, 2018 reply Follow Share @puja :unflag tthe answer, i have corrected it now ! 0 votes 0 votes Puja Mishra commented Jan 30, 2018 reply Follow Share I hav flagged it because that editor named pawan singh .... that may be his name ... he edited things wrongly in many questions and answers ... 0 votes 0 votes akshay_123 commented Sep 4, 2023 reply Follow Share wrong and edited 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes This question can be done simply as question is asking for at least one of them facing 6 .. P(at least one 6 face up)= 1 - (none of them facing 6) 1 - ( 5/6 * 5/6) 1- 25/36 11/36.. Sandeep Suri answered Jan 8, 2018 Sandeep Suri comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes P(atleast one of dice will have 6 facing) = 1 - P(none of dice have 6 facing up) = 1 - ( 5/6 * 5/6) =1- 25/36 = 11/36 akshay_123 answered Sep 4, 2023 akshay_123 comment Share Follow See all 0 reply Please log in or register to add a comment.