From a deck of five cards numbered 2, 4, 6, 8, and 10, respectively, a card is drawn at random and replaced. this is done three times. what is the probability that the card numbered 2 was drawn exactly two times, given that the sum of the numbers on the three draws is 12?

Since the sum of the numbers on the three draws is 12, if we want the card numbered 2 to be drawn exactly two times, the third card can only be numbered 8. In fact, , and there are no other possibilities, unless you consider the various permutations of the terms.

So, we have three favourable cases: we can draw 2,2,8, or 2,8,2, or 8,2,2. This are the only three cases where the card numbered 2 is drawn exactly two times, and the sum of the number on the three draws is 12.

Now, the question is: we have three favourable cases over how many? Well, we have 5 possible outcomes with each draws, and the three draws are identical, because we replace the card we draw every time.

So, we have 5 possible outcomes for the first draw, 5 for the second and 5 for the third. This leads to a total of possible triplets.

Once we know the "good" cases and the total number of possible cases, the probability is simply computed as

-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

answer: '

the distance between points l and m

step-by-step explanation: