D DDSelin2mori Full Member 10+ Year Member 15+ Year Member Joined Jul 11, 2007 Messages 66 Reaction score 0 Points 0 Pre-Dental Apr 28, 2008 #1 Advertisement - Members don't see this ad With one fair die what is the probability of having two fours in five attempt? I'm solving in this way 5C2/6^5 which is wrong!! why this is wrong? and how do you solve it? thanks
Advertisement - Members don't see this ad With one fair die what is the probability of having two fours in five attempt? I'm solving in this way 5C2/6^5 which is wrong!! why this is wrong? and how do you solve it? thanks
S Streetwolf Ultra Senior Member Verified Member 10+ Year Member Dentist 15+ Year Member Joined Oct 25, 2006 Messages 1,801 Reaction score 7 Points 4,571 Location NJ Dentist Apr 28, 2008 #2 (5C2) chooses the rolls which are 4s. (1/6)^2 describes the rolls which are 4s. (5/6)^3 describes the other rolls which are not 4s. So you have (5C2)*(1/6)^2*(5/6)^3. Your answer needs to be multiplied by 5^3 = 125. Upvote 0 Downvote
(5C2) chooses the rolls which are 4s. (1/6)^2 describes the rolls which are 4s. (5/6)^3 describes the other rolls which are not 4s. So you have (5C2)*(1/6)^2*(5/6)^3. Your answer needs to be multiplied by 5^3 = 125.
