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I'm not that good....just sometimes lucky.....

(a)

Prob 0 green  =  5C3 / 9C3  = .12

Prob  1 green  =  3C1 * (4/9)(5/9)^2  = .41

Prob 2 green 3C2 * (4/9)^2 * (5/9)  = .33

Prob 3 green  = (4/9)^3  = .09

The last histogram is correct

(b)

Expected number of green balls

0*(.12) + 1*(.41)  + 2*(.33) + 3*(.09)  = 1.34

Thanks! You are a math genius bro. How you get so good at math?

16%  of the population  will be 1 standard deviation (or more)  below the mean.....this is a score of 75

So....this represents all the scores below 75

[ 16% will also have scores above 85....but this isn't a choice ]

Look at the columns under f(x) and g(x)   that have equal values for both

Note these occur  at   x  = 0 , x = 1  and x = 2

cos x = .873

Take the arccos to find x

So

arccos (.873)  = x  ≈  29.2°

The other angle will be  [ 360 - 29.2]  ≈ 330.8°

The purple figure is a para\(\)llelogram, with an area of 5*6=30, since base*height=area...

The next figure is a trapezoid...and it's area is  \(\frac{1}{2}{10}{h}=30\), and 5h=30, h=6.

-From Sneaky, maybe CPhill can guess who I am...

Area of quadrilateral  =  6 * 5  = 30

Area of trapezoid  =  (1/2) h  * ( sum of base lengths )

30 = (1/2) h ( 3 + 7)

30  = (1/2) h (10)

30  =  5h    divide both sides by  5

6  =  h

Since DX is perpendicular to base BC....it  forms a height of the parallelogram...

Area  =  12 * DX

144 = 12 * DX     divide both sides by 12

12 = DX

16% of the population has z-scores from -infinity to -1

Thanks!!!

Times a 7 is rolled         Prob 

      0                           ( 6/7)^4  ≈    .5398

      1                          C(4,1) (6/7)^3 (1/7)  ≈ .3599

      2                          C(4,2) (6/7)^2(1/7)^2  ≈  .09

      3                          C(4,3) (6/7)(1/7)^3  ≈ .001

      4                           (1/7)^4  ≈   0

thank you

0  green balls  = (3/7)^3  ≈ .079

1 green ball  =  C(3,1) (4/7)*(3/7)^2  = .315

2 green balls = C(3,2) (4/7)^2*(3/7)  = .42

3 green balls  =  (4/7)^3  = .187

Total number of committees  = 11C3  = 165

Committees with :

0 women on the committee  =  5C3  = 10

1 woman on the committee = 6C1 * 5C2  =  60

2 women on the committee = 6C2 * 5 1  = 75

3 women on committee  = 6C3  = 20

Expected number of women =

0(10/165) + 1(60/165) + 2(75/165) + 3(20/165)  ≈ 1.64

Total possible committees  = 30C3  = 4060

Committees with 0 juniors  =  19C3  = 969

Committes with 1 junior  = 11C1 * 19C2  = 1881

Committees with 2 juniors  = 11C2 * 19C1  = 1045

Committees with 3 juniors  = 11C3  = 165

So...the expected value is

0(969/4060) + 1(1881/4060) + 2(1045/4060) + 3(165/4060) =  1.1

One question,Rom...[I'm not too familiar with this....]

Don't we have

(0)(57/203) + 1(95/203) + 2(45/203) + 3(6/203)   by  the summation formula???

Times a 7 is rolled         Prob 

      0                           ( 6/7)^4  ≈    .54

      1                          C(4,1) (6/7)^3 (1/7)  ≈ .36

      2                          C(4,2) (6/7)^2(1/7)^2  ≈  .09

      3                          C(4,3) (6/7)(1/7)^3  ≈ .001

      4                           (1/7)^4  ≈   0

We have   2^5  =  32 possible outcomes

0  heads  =  1 /32

1 head  =  5/32

2 heads  = 5C2/32  = 10/32  =  5/16

3heads  = 5C3/32  = 10/32  =  5/16

4 heads = 5/32

5 heads =1 /32

\( (\sqrt{2}+\sqrt{3})^3\)

\(a\sqrt{2} + b\sqrt{3} + c\sqrt{6} \)

We just have a binomial expansion  .....

( √2)^3  +  3(√2)^2(√3)  + 3(√2)(√3)^2  + (√3)^3       simplify

2√2 + 6√3  + 9√2  + 3√3  =

11√2 + 9√3 + 0√6

So.....a + b + c  =   20

wow, I totally messed this one up... have to stop answering these when I'm drinking! 

Let's start again shall we

\(P[k] = \dfrac{\dbinom{20}{k}\dbinom{10}{3-k}}{\dbinom{30}{3}}\\ P[(0,1,2,3)] = \large \left(\frac{6}{203},\frac{45}{203},\frac{95}{203},\frac{57}{203}\right)\)

\(E[X] = 0\cdot \dfrac{6}{203} + 1 \cdot \dfrac{45}{203}+2\cdot \dfrac{95}{203}+3\cdot \dfrac{57}{203} =2\)

Total  number of possible committees   =  30C3  = 4060

Number of committees  with 0  junior members  = 12C3  = 220

Number of committees with 1 junior member and 2 senior members = 18C1 * 12C2  = 1188

Number of committes with 2 junior members and 1 senior member = 18C2 * 12C1  = 1836

Number of committees with 3 junior members  = 18C3  = 816

Expected number of juniors  =

0*(220/4060) + 1* (1188/4060)  + 2* ( 1836/4080) + 3 *(816/4060) ≈  1.8

If it is a fair die  7 should be rolled 1 time out of 7   or 1/7 of the time    1/7 x 4 rolls =  4/ 7

Juniors make up 20 out of 30 of the members   or 2/3     

You would expect 2/3 of the committee to be juniors   2/3 x 3 = 2 committee members should be juniors