Two number cubes are rolled --one white and one yellow. How do you find the probability that the white cube shows a 6 and the sum is greater than 9?

Question

Genesis Allen · Accepted Answer

#P("white cube" = 6, "sum of both cubes" >9)=3/36=1/12#

I'm going to assume that we're working with fair, standard, 6-sided dice.

Let's look at the possible rolls (I'll highlight the white cube's 6 and sums greater than 9 with #color(blue)"blue"#):

#((color(white)(0),ul1,ul2,ul3,ul4,ul5,ul6),(1|,2,3,4,5,6,7),(2|,3,4,5,6,7,8),(3|,4,5,6,7,8,9),(4|,5,6,7,8,9,color(blue)10),(5|,6,7,8,9,10,color(blue)11),(6|,7,8,9,10,11,color(blue)12))#

There are 3 ways we can have the white cube be a 6 and the sum of the two cubes be greater than 9. This is out of 36 possible rolls, and so we can say:

#P("white cube" = 6, "sum of both cubes" >9)=3/36=1/12#

Isaiah Anthony · Answer

undefined To find the probability that the white cube shows a 6 and the sum is greater than 9, we first determine the outcomes where the white cube shows a 6 and the sum is greater than 9. Since the maximum value of a standard six-sided cube is 6, the only way the sum can be greater than 9 is if the yellow cube also shows a 4, 5, or 6.

So, the favorable outcomes are when the white cube shows a 6 and the yellow cube shows a 4, 5, or 6. There are 3 favorable outcomes out of a total of 6 possible outcomes for the yellow cube (1, 2, 3, 4, 5, 6).

Therefore, the probability is 3 out of 6, which simplifies to 1 out of 2, or 0.5.