Melody

It is because the other 2 posts have been blocked.

Sometimes this happens automatically and sometimes a moderator will block it.

Sometimes posts get auto blocked for no good reason.

If this happens you can write a private message to a moderator and they can unblock it for you.

You will need to include the address of the post, which I think you will still have access too.

You will obviously have to wait till they read your message.

Also old questions get locked (not blocked) automatically.

If you want one to be opened temporarily so you can add a post just request it of a moderator.    

A bag contains five white balls and four black balls. Your goal is to draw two black balls.

You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

Winning draws

BB                        (4/9)*(3/8)   =  12/72

BW   B                 (4/9)*(5/8)*(3/8)    =  60/576

WB    B                 (5/9)*(4/8)*(3/8)    =  60/576

WW   BB               (5/9)*(4/8)*(4/9)*(3/8)  =  240/5184

ADDED = 12/72+120/576+240/5184 = 91/216

Non winning draws

WW   WB             (5/9)*(4/8)*(5/9)*(4/8)    =   400/5184

WW   BW             (5/9)*(4/8)*(4/9)*(5/8)      =   400/5184

WW   WW             (5/9)*(4/8)*(5/9)*(4/8)    =    400/5184

WB     W                (5/9)*(4/8)*(5/8)            =    100/576

BW      W                (4/9)*(5/8)*(5/8)            =   100/576

ADDED= 1200/5184+200/576 =125/216

125+91=216 great.

I draw a probability tree and listed all the probabilities on it.

I ended up with 9 possible scenarios and 4 of them were favourable.

If I work out the probabilities of each of the favourable ones and add them together then hopefully that will be the answer.

Although I would check it by adding all possible outcomes probabilities together (They should add to 1)

Attn: Iamhappy

I do not like that you have given your guest answerer thumbs down.

If someone answers you, you should thank them for their interest and their time.

If you do not think the answer is correct then you open a polite conversation with your reasons.

To give thumbs down to someone who has tried to help you is rude.

Note: I have not looked at the answer as yet so I have no idea whether or not it has any merit.

But thank you guest answerer for your time and interest.

Determine f(6)

Just substitute 6 for x in the equations

f (x) = - 2/3 x - 1

f (6) = - 2/3 *6 - 1 = -4-1 =-5    So one point on the graph is (6,-5)

Now find f(-9)

Determine x if f(x) = –7

this time it is the other way around

You have to solve

-7=-2/3x-1      

That is a good start for you.

(5x^4 -8x^3 +2x^2 +4x+7)/(x+2)^4 = a+b/(x+2) +c/(x+2)^2 +d/(x+2)^3 +e/(x+2)^4

\(\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =a+\frac{ b}{(x+2)} +\frac{c}{(x+2)^2} +\frac{d}{(x+2)^3} +\frac{e}{(x+2)^4}\\ \\~\\ \frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =\frac{a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e}{(x+2)^4}\\~\\ (5x^4 -8x^3 +2x^2 +4x+7) =a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\ (5x^4 -8x^3 +2x^2 +4x+7) =a(x^4+8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\ a=5\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+3*2*x^2+3*4x+8)+c(x+2)^2+d(x+2) +e\\ ( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\ -8=40+b\\ b=-48\\~\\ ( 2x^2 +4x+7) =5(24x^2+32x+16)+ -48(6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\ \)

Just keep going till you have them all.

There is probably a better way though.

Coding:

\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =a+\frac{ b}{(x+2)} +\frac{c}{(x+2)^2} +\frac{d}{(x+2)^3} +\frac{e}{(x+2)^4}\\
\\~\\
\frac{(5x^4 -8x^3 +2x^2 +4x+7)}{(x+2)^4} =\frac{a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e}{(x+2)^4}\\~\\
(5x^4 -8x^3 +2x^2 +4x+7) =a(x+2)^4+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\
(5x^4 -8x^3 +2x^2 +4x+7) =a(x^4+8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\
a=5\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x+2)^3 +c(x+2)^2+d(x+2) +e\\~\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+3*2*x^2+3*4x+8)+c(x+2)^2+d(x+2) +e\\
( -8x^3 +2x^2 +4x+7) =5(8x^3+24x^2+32x+16)+ b(x^3+6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\
-8=40+b\\
b=-48\\~\\
( 2x^2 +4x+7) =5(24x^2+32x+16)+ -48(6x^2+12x+8) +c(x+2)^2+d(x+2) +e\\

Thanks very much Heureka :)

I would start by drawing a probability tree.  Do you know how to draw one?

Thanks very much for the explanation.

I do not remember seeing such a question done like this before.

You have taught me, I hope I retain the knowledge.