Well for i) Probability of winning first dart = p
Probability of winning with second dart given first dart did not hit
(1-p) )
If you expand that you should get:
For ii) To win, she needs to hit the target twice, there are
4 possibilities:
There are
3 possibilities of hitting 2 darts with one missing (i.e. 1st and 2nd, 2nd and 3rd, 1st and 3rd).
The probability of this is:
 )
1 permutation of hitting with all 3 darts:
So the total probability
 + p^3 = 3p^2 – 2p^3 )
For iii) I hope I didn't type anything wrong...
: data from (i) $)
: data from (ii) $)
If Darcy is to be more likely to win game 1 than game 2, then from above:
 > p (3p – 2p^2))
Since
^2 > 0)
Therefore for 0 < p < 1, this statement is true and the induction holds.
For iv), algebra can really finally be used to some extent.
 )
Divide by p, get it into the format of
and use the quadratic formula:
Thus, you have a quadratic equation to solve and the quadratic formula is used which I cannot bother with typing unfortunately...
Eventually you get p = 1.39, 0.36. Only one of these values is between our limit 0 < p < 1, so the answer would be when p = 0.36 (2 dp).
but ive gotta get my swimming up, sorry 
