A jar contains only one blue tile, one green tile, one red tile, and one yellow tile. If a tile is chosen at random and then replaced in the jar, and then a second tile is chosen at random, how many of the possible outcomes contain at least one red tile?
1
2
4
7

7

Step-by-step explanation:

Since there are 4 options for the first pick and 4 options for the second, there are a total of 4*4=16 outcomes. Out of these, if the first tile is red, then there are four tiles that the second could be. Similarly, if the second tile is red, there are four tiles the first could be. Together this is 8, but this includes the overlap where both tiles are red, which means that you are counting twice. Therefore, the answer is 8-1=7. Hope this helps!

C. 7

Step-by-step explanation:

Let's do some casework here.

Case 1: You only get 1 red tile

If you get one red tile, then that means we don't want the second tile to be red again. So, we can have the six possible pairings:

RG

RB

RY

GR

BR

YR

Note that order matters. Here, R denotes a red tile, G denotes a green tile, B denotes a blue tile, and Y denotes a yellow tile.

Case 2: You get 2 red tiles

If we get 2 red tiles, the only possible pairing is RR.

So, we have 6 possibilities from Case 1 and 1 possibility from Case 2, so add these together to get 1 + 6 = 7 possible outcomes.

The answer is D.

0

step-by-step explanation:

you can not divide by 0 so 4 ÷ 0 would be 0

0 × 9999 =0

mark brainliest   : )

step-by-step explanation: