a child rolls a 6-sided die 6 times. what is the probability of the child rolling no more than three twos g

Pr(Three 2's) = 1/27

Step-by-step explanation:

Let's assume the die is a fair die, on the first roll of the die, the child has a 1/6 chance of getting any number, including 6.

Second roll, the child has a 1/36 chance of getting any two numbers, including two 6's.

And on the third roll, the child has a 1/36×1/6=1/216 chance of getting any three numbers, including three 6's. And this is due to the fact that the rolls are independent, so the total possible outcomes multiply each roll with each roll's probability. Since each roll's probability is 1/6.

The probability of the child rolling no more than three twos will be =2/6×2/6×2/6

=1/3×1/3×1/3

=1/27

Therefore, the chances of three twos will be 1/27

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

step-by-step explanation: