algebra tree diagrams

Question 1

Example 1 -

There are 5 red counters and y blue counters in a bag

Imogen takes a counter from the bag at random
She puts the counter back into the bag
Imogen then takes another counter at random from the bag

The probability that the first counter imogen takes is red and the second counter Imogen takes is red is 1/9

Work out how many blue counters are in the bag

Answer 1

Red: 5 Blue: y Total: y+5

               Red: 5/y+5 Red: 5/y+5
              Blue: y/y+5

              Red: 5/y+5 B: y/y+5 
              Blue: y/y+5

P(RR)=1/9
5/y+5 x  5/y+5 = 1/9
25/(y+5)(y+5) = 1/9
Multiply both sides by (y+5)(y+5)
25=1/9(y+5)(y+5)
multiply both sides by 9
225 = (y+5)(y+5)
Expand the brackets
225=y^2+5y+5y+25
simplify
225=y^2+10y+25
Make it equal zero by minusing 225
y^2+10y-200=0
Now we use the ac method
A number that:
multiplies to give -200
Adds to give 10
20 and -10 give this
(y+20)(y-10)=0

If:
y+20=0
y=-20
This isn’t possible as there can’t be a minus number of counters

If:
y-10=0
y=10
This is possible so there are 10 blue counters

Question 2

Example 2 -

There are 5 red counters and x blue counters in a bag
2 counters are removed from the bag at random
The probability that both counters taken are red is 5/33
Work out the value of x

Answer 2

Blue: 5 Red: x Total: x+5

               Red: 4/x+4 Red: 5/x+5
              Blue: x/x+5

               Red: 5/x+4 Blue: x/x+5
               Blue: x-1/x+4

P(RR)=5/33
5/x+5 x 4/x+4 = 5/33
20/(x+5)(x+4)  = 5/33
expand the brackets
20/x^2+9x+20 = 5/33
multiply both sides by the denominator
20= 5/33 (x^2+9x+20)
multiply by 33
660=5(x^2+9x+20)
Expand the brackets
660=5x^2+45x+100
Make it equal zero by minusing 660
5x^2+45x-560=0
divide by 5
x^2+9x-112=0
Need two numbers that:
multiplies to give -112
adds to give 9
16 and -7
(x+16)(x-7)=0

If: x+16=0
x=-16
Not possible

If: x-7=0
x=7
Possible

There are 7 blue counters in the bag

Question 3

Example 3 -

There are 4 red counters and x blue counters in a bag.
2 counters are removed from the bag at random
The probability that both counters taken are blue is 1/3
Work out the value of x

Answer 3

Red: 4 Blue: x Total: x+4

             Red: 3/x+3 Red: 4/x+4
             Blue: x/x+3

              Red: 4/x+3 Blue: x/x+4
              Blue: x-1/x+3

P(BB)=1/3
x/x+4 x x-1/x+3 = 1/3
x(x-1)/(x+4)(x+3) = 1/3
multiply by the denominator
x(x-1)=1/3(x+4)(x+3)
multiply by 3 and expand the brackets
3x^2-3x=x^2+4x+3x+12
Simplify
3x^2-3x=x^2+7x+12
minus x^2+7x+12
2x^2-10x-12=0
divide by 2
x^2-5x-6=0
Two numbers that:
multiply to give -6
Add to give -5
-6 and 1
(x-6)(x+1)=0

If:
x-6=0
x=6
Possible

If:
x+1=0
x=-1
Impossible

There are 6 blue counters

Question 4

Example 4 - Half example

There are some red counters and some blue counters in a bag
The ratio of red counters to blue counters is 3:1
Two counters are removed at random
The probability that both counters taken are blue is 1/20
Work out how many counters were in the bag before any counters were removed

              Red: 3x-1/4x-1 Red: 3x/4x
             Blue: x/4x-1

           Red: 3x/4x-1 Blue:x/4x
           Blue: x-1/x-1

Answer 4

Red: 3x-1/4x-1
Red: 3x/4x
Blue: x/4x-1

           Red: 3x/4x-1 Blue:x/4x
           Blue: x-1/x-1

Won’t finish the question off but this is what your tree diagram should look like