This is a rather famous probability problem.
The easiest way to solve this is to calculate the probability that you WON'T roll a "double 6" (or a twelve) each time you roll the dice. There are 36 ways in which dice rols can appear and only one is a twelve. So, for one roll, the probability that you will NOT get a twelve is (35/36)^n where 35/36 is about .97222222 and n would equal 1 for the first trial. So for your first roll the odds that you WON'T get a 12 is .97222222.
For the second roll we calculate (35/36) to the second power or (35/36)^2 which equals about .945216.
When we get to the 24th roll we calculate (.97222222)^24 which equals 0.508596.
For the 25th roll, we calculate (.97222222)^25 which equals 0.494468. For the first time we have reached a probability which is lower than 50 per cent. That is to say, after 25 rolls, we have reached a point in which the probability is less than 50 per cent that we will NOT roll a twelve.
To phrase this more clearly, after 25 rolls we reach a point where the probability is greater then 50 per cent that you will roll a 12 at least once.
Please go to this page 1728.com/puzzle3.htm and look at puzzle 48. (The last puzzle on the page). An intersting story associated with this probability problem is that in 1952, a gambler named Fat the Butch bet someone $1,000 that he could roll a 12 after 21 throws. (He miscalculated the odds [as we know you need 25 throws] and after several HOURS, he lost $49,000!!!)
Please go that page and it has a link to the Fat the Butch story.
Awnser is below in the photo
Answer:
C- Divide both sides by 3.
Step-by-step explanation:
In order to isolate the variable, we will have to remove all non-1 coefficients from the variable. Since 3b can be written as 3(b), b is being multiplied by 3. So the only way to isolate the variable is to do inverse operations, which would be to divide both sides of the inequality by 3.
Hope this helped!
this picture would help you man stay strong
