Question 939371
  
Given: a bag of 2 hamburgers (H), 4 cheeseburgers (C), and 3 chicken burgers (K).

Solution
(a) A sandwich is randomly selected from the bag. Find probability of picking one of Wyatt's order -- either a cheeseburger or chicken.
Assuming random selection, the probability of selecting any burger has equal probability and the probability of success is the ratio #success/#possibilities.
Success is choosing one of 4C or 3K with a total of 7.  The bag contains 9 burgers, which means
P(C or K) = {{{7/9}}}
  
(b) Two A sandwiches are randomly selected from the bag (assumed without replacement).  Find probability of not picking Wyatt's order in both picks.
The only choice which is not in Wyatt's order is hamburger (H).
There are two hamburgers in the bag.  The first pick is 2 out of 9, and the second pick is 1 (remaining) out of 9.  Using the multiplication rule for a two step experiment (without replacement), the probability of picking none of Wyatt's order in the first two picks is
{{{P(HH)=(2/9)(1/9)=2/81}}}
  
Answer:
(a) probability of picking one of Wyatt's choice is 7/9.
(b) probability of not picking Wyatt's choice in both of two picks is 2/81.