Please help asap >.<

? what 

Part 1 involving the Spinner

Firstly, let's look at the first probability statement. We have an ample chance of landing the same spin again, so we won't be removing one blue for the next spin.

This fits the denominator 64, as that is 8*8. 

We first have a 5/8 chance to land on a blue, then we have a 3/8 chance to land on a red. 

\({5\over{8}}*{3\over{8}}\)is our equation (sorry for it being tiny, LaTeX is small)

In fraction multiplication, we multiply the top and bottom separately. 5*3 is 15, and as we said before 8*8 is 64

This leaves us with the fraction 15/64, which is true.

Finding this answer ALSO eliminates the third answer from being true, as it said that we remove a piece of the spinner if it is landed on, which is untrue.

Keep in mind that (red, blue) is the same as (blue, red)

-------------------------------------------------------------------------------

Now, let's look at the second option. 

The probability of (blue, blue) is \({5\over{8}}*{5\over{8}}\)

Now, let's multiply. 5*5 is 25, and 8*8 is 64. 

Therefore, this answer is true. 

-------------------------------------------------------------------------------

Finally, let's look at the fourth option. (Remember that the third option was not correct.)

The probability of (red, red) is \({3\over{8}}*{3\over{8}}\)

Now, let's multiply. 3*3 is 9, and 8*8 is 64.

9/64 does NOT reduce to 3/32. 

Therefore, this is a FALSE statement. 

The correct statements are Options 1 and 2.

-------------------------------------------------------------------------------

Part 2 involving Playing Cards

First, let's look at the part involving (Queen, Black Card)

First, this is a 4 in 52 or 1 in 13 chance of drawing a queen.

Then we draw a black card. 

The chance of this is 1 in 2.

We'll use the simplified numbers to form our equation.

\({1\over{13}}*{1\over{2}}\)

Now we multiply. 1*1=1, and 13*2=26

We now end with 1/26. 

-------------------------------------------------------------------------------

Now let's look at the probability of (Diamond, Ace)

First, the chance of drawing a Diamond in a set of 52 cards is 13 out of 52, or 1 in 4.

Then you have a 4 in 52 or 1 in 13 chance of drawing an Ace after the card is replaced.

Let's set up the equation: \({1\over{13}}*{1\over{4}} \)

Now, let's multiply. 1*1=1, and 13*4=52.

Therefore, our answer is 1/52.

-------------------------------------------------------------------------------

Thirdly, let's look at the probability of (Jack, 4)

First, the probability of drawing a Jack is 1 in 13.

Then, after replacement, the probability of drawing a 4 is 1 in 13.

We can make this easy on ourselves by squaring the fraction since they are the same. \({1\over{13}}^2\)

1^2=1, and 13^2=169

Therefore, our final answer is 1/169.

-------------------------------------------------------------------------------

Finally, let's look at the probability that we first draw a red card, and then a Club (Red card, Club)

The probability of drawing a red card is 1 in 2.

The probability of drawing a club is 1 in 4.

Now, let's set up the equation: \({1\over{2}}*{1\over{4}}\)

Now let's multiply. 1*1=1, and 2*4=8

Therefore, our answer is 1/8.

-------------------------------------------------------------------------------

Hope this helped and sorry for the long answer!