Solution for Section 6.3 Question 2

2.a) Your sample space, S, is all the ways of choosing three dishes from the take away. There are 27 possible dishes to choose from so  

n(S) = 27 · 27 · 27 = 19683.

Consider the event, E,  that you all choose chicken dishes,  then

n(E) = 6 · 6 · 6 = 216.

The probablility that all three of you choose  chicken dishes is n(E) / n(S) = 216 / 19683 = 8 / 729. Hence the probablility that the three of you end up with a meal that does not consist only of chicken dishes is

1 - 8 / 729 = 721 / 729.



b) Your sample space is the same as in part a). Here let E be the event that none of you choose vegetarian dishes. Ignoring the vegetarian dishes, you each have 22 dishes to choose from, so

n(E) = 22 · 22 · 22 = 10648.

The probability that the three of you end up with a meal that has no vegetarian dishes in it is

P(E) = n(E) / n(S) = 10648 / 19683.

Back to Section 6.3