@SolomonZelman

Question

@SolomonZelman

anonymous · Answer

hold on.

anonymous · Answer

1. random 
2. 32
3. A.

solomonzelman · Answer

I think you can do number one yourself :) take a shot please:)

anonymous · Answer

i just asked you to check my answers......

solomonzelman · Answer

What are your answers ?

anonymous · Answer

i posted them above the picture.

solomonzelman · Answer

OK, I think that the 1st 2 are correct, I don't see the options for number 3 though.

anonymous · Answer

attachment

solomonzelman · Answer

Number 3.
P(n,r) =  P(15,15) =
15! / (15 - 15)! = 
15! / (0)! = 
15! / 1 =
15! = 
$$1 \times 2 \times 3 \times 4 \times 5 ... \times 14 \times 15$$
1,307,674,368,000

solomonzelman · Answer

You are correct for every one of the answers :)

anonymous · Answer

4. 35
5. 1/3

solomonzelman · Answer

4.  Correct   (nCr function)

5.   Correct  (2/6=1/3)

solomonzelman · Answer

You are smart !!

anonymous · Answer

6. 12/17

anonymous · Answer

7. 2.1
8. 2/9

solomonzelman · Answer

6. Correct (  40+50  /  40+80+50  = 120 / 170 = 12/17  )

solomonzelman · Answer

I am quite not getting the question, sorry-:(
well, if you are asking for the probability of the spinner landing on the dark, out of the all sections, then it is 2:3 (b/c 2:3 is same as 2/3 )

If you mean the proportion of the times it can land on the dark, vs the time it can land on the light, then it is 2:1

solomonzelman · Answer

For number 8.

11=6+5
7=6+1 , 5+2, 4+3

the rest is 
1+1
1+2
1+3, 2+2
1+4, 2+3
1+5, 2+4, 3+3  (skipping the sum of 7)
2+6, 3+5, 4+4
3+6, 4+5
4+6, 5+5
6+6

so all the sums possible including 11 and 7, are
21 sums

and there are 4 sums that can be 11 and 7,

So 4/21 (I am not 100% sure on this one)

anonymous · Answer

that's not one of the answer choices.