3 votes 3 votes One dice is thrown three times and the sum of the thrown numbers is 15.find the probability for which number 4 appears in first throw. Probability engineering-mathematics probability conditional-probability + – indrajeet asked Aug 6, 2016 • recategorized Sep 15, 2016 by sourav. indrajeet 1.2k views answer comment Share Follow See 1 comment See all 1 1 comment reply vijaycs commented Aug 7, 2016 reply Follow Share 1/18 ?? 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes Total no of patterns where 15 as sum appears... 366,456,465,546,564,555,645,654,663,636 So probability = 2/10 = 1/5 papesh answered Aug 7, 2016 • edited Aug 7, 2016 by papesh papesh comment Share Follow See all 4 Comments See all 4 4 Comments reply indrajeet commented Aug 7, 2016 reply Follow Share I am getting same but ans is 1/18 0 votes 0 votes papesh commented Aug 7, 2016 reply Follow Share Question already includes that sum is 15 ... We need to find only how may cases it will give 15..among these 15 how many are favourable... That gives the probability.. 0 votes 0 votes cse23 commented Aug 10, 2016 reply Follow Share @gabbar n(s) = 6*6*6 =216 it is already given that 4 appears in 1st throw so in 2nd and 3rd throw we can have (5,6) and (6,5) to make sum as 15 so n(E) ={(4,5,6),(4,6,5)} p(E)=2/216 =1/105............ryt?? –1 votes –1 votes Arnab Bhadra commented Jun 15, 2017 reply Follow Share Here Sample Space is 10 (sum of the thrown number is 15). 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes its 2/10 ie 1/5 Manikant Kumar answered Sep 13, 2016 Manikant Kumar comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes ans = P( number in 1st dice throw is 4 & sum of numbers in 2nd and 3rd dice throw is 11 ) / P(number in 1st dice throw is 4) = (1/6 * 2/36 ) / ( 1/6 ) =1/18 Sushant Gokhale answered Sep 9, 2016 Sushant Gokhale comment Share Follow See all 2 Comments See all 2 2 Comments reply Injila commented Aug 12, 2017 reply Follow Share Why you have divided by P(no. in 1st dice throw is 4). As condition is given as sum is 15, and we divide with the probability of given condition in conditional probability, right? 0 votes 0 votes Sushant Gokhale commented Aug 12, 2017 reply Follow Share I am sorry. It should be Ans= p (4 on first dice and sum of other 2 dice is 11) /{ p (4 on first dice and sum of other 2 dice is 11)+ p (First dice has other than 4 and sum of all 3 is 15)} =2/10 =1/5 0 votes 0 votes Please log in or register to add a comment.