Question 547249
We need to know how Paul wins to answer this question, but let's just assume he wins if it lands on red, and hopefully this helps.


Since the wheel is equally likely to land on any of the 4 areas, the probability that it lands on red is {{{1/4}}} (as is the probability of each of the other areas).

The probability that Paul wins 3 times is 
{{{(1/4)*(1/4)*(1/4)*(3/4)*(3/4)=9/1024}}}
(Note: we are using the product rule (the "or" rule), and also note that {{{3/4}}} is the probability that the wheel will not land on red).

Similarly,
 
The probability that Paul wins 2 times is 
{{{(1/4)*(1/4)*(3/4)*(3/4)*(3/4)=27/1024}}}


The probability that Paul wins 1 times is 
{{{(1/4)*(3/4)*(3/4)*(3/4)*(3/4)=81/1024}}}

And 

The probability that Paul wins  times is 
{{{(3/4)*(3/4)*(3/4)*(3/4)*(3/4)=243/1024}}}


Now the probability that Paul wins 3 or fewer times we can get by using the sum rule and adding each of these individual probabilities.
{{{9/1024+27/1024+81/1024+243/1024=360/1024=45/128}}}