why is this approach wrong?
probablility=probability of squash x probability of peas x probablity of fruit x probability of salad
You don't multiply each probability separately...
the probability of choosing squash and peas is NOT 1/3 x 1/3 = 1/9 because the patrons can choose "two from a group of three." So you need to calculate the "combination".
So the prob of choosing squash and peas from the vegetables group is
3C2 = 3.
This is also the case for choosing two out of three in the appetizer group.
3C2 = 3
So total possible combinations are 3 x 3 = 9.
Therefore, 1/9. B (as is the OA)