Help on maths hw tree diagram

can someone tell me how to figure this out please!?

line a makes bags containing 10 chocolates

each Chocolate has either a soft or hard center

2 random chocolates are chosen at random from the bag

the probability that both chocolates are soft is 30/90

complete the tree diagram (then its a empty tree diagram)


HELP PLEASE SOMEONE! I GOT A EXAM TOMOZ AND MY SOUL IS SLOWLY LEAVING MY BODY WITH THIS Q!

It's impossible drawing a tree diagramm here so I'm just going to explain.
Judging from the final probability of 30/90, the 2 chocolates are picked without replacement as the denominator is 10 x 9.

The tree diagram has 2 lots of branching (probabilities change on 2nd branch because there is no replacement):
if 1st chocolate: Soft [x/10] 2nd chocolate: Soft [(x-1)/9)] Hard [(10-x))/9
if 1st chocolate: Hard [(10-x)/10] 2nd chocolate: Soft [x/9] Hard [(10-x-1)/9]


The probabilities are shown in the sqaure brackets, x denoting the number of soft chocolates initially.
To get 2 soft chocolates, must form the equation:
(x/10)[(x-1)/9] = 30/90
so [x(x-1)]/90 = 30/90
so x must be 6 and x-1 must be 5

From this we know that there were 6 soft chocolates in the beginning. Now just substitute x into the tree diagram with the unknown probablities. So now you have the answer:

if 1st chocolate: Soft [6/10] 2nd chocolate: Soft [5/9)] Hard [4/9
if 1st chocolate: Hard [4/10] 2nd chocolate: Soft [6/9] Hard [3/9]


Hope I helped :D

S= soft, H= hard

X=number of soft

P(SS)= (30/90)= (x/10)(x-1)/9=x(x-1)/90
X(x-1)=30
x=6 because 6*5=30, but you could solve it as a quadratic.

So 6 are soft and 4 are hard. You are starting out with 6S and 4H
Then P(S)=6/10 and P(H)=4/10 for the first candy.
Then you draw a second one:
If the first was hard, that leaves 3 out of 9 that are hard, and 6/9 soft.
If the first was soft, that leaves 5/9 soft and 4/9 hard

Okay, i will try to draw the tree:)


..H(4/10)....................... ....S(6/10).......first candy drawn
........^................ ....................^
S(6/9).. H.(3/9) ..............S.(5/9))..... H(4/9).......second candy (one is already gone)
...v............v......... ..... ........v........ ......v...
HS.........HH....... ...........SS....... ........SH...these are the possible outcomes
(4/10)(6/9)..(4/10)(3/9).....(6/10)(5/…

24/90..............12/90 ..........30/90 ...........24/90

8/30............4/30.............. 1/3...............8/30

Hoping this helps!