Question

guests at a dinner event can choose 1 of 4 appetizers, 1 of 3 main dishes, and 1 of 6 desserts. jay and kim are seated next to each other, and learning they ordered the same dinner combination, decide it must be fate that they should be best friends. what was the probability that jay and kim would have chosen the same dinner combination if both chose randomly?

Explanation:

Step1: Calculate total combinations

The total number of ways to choose an appetizer, main dish, and dessert is \(4\times3\times6 = 72\) (using the multiplication principle: \(n_{total}=n_{appetizer}\times n_{main}\times n_{dessert}\)).

Step2: Calculate probability

Since Jay and Kim choose randomly, the probability that they choose the same combination is \(\frac{1}{72}\) (because for any combination Jay picks, Kim has a \(1\) in \(72\) chance to pick the same one).

Answer:

\(\frac{1}{72}\)