Tad draws three cards at random, without replacement, from a deck of ten cards numbered

1 through 10. What is the probability that no two of the cards drawn have numbers that
differ by 1? Express your answer as a common fraction.

there are 9 pairs of cards that differ by one

... 1-2, 2-3, 3-4, etc.

there are 10C2 possible pairs ... 45

so the probability of a pair differing by one is 9/45 or .2
... by more than one is (1 - .2) = .8

3 cards contain 3 pairs

.8^3 = .512

Thanks a lot Scott, Appreciate your quick response.

I reviewed the answer key and it states 7/15 as the answer to this problem. I am wondering if the answer key is incorrect.

Thanks again
Nikhil

To find the probability that no two cards drawn have numbers that differ by 1, we can count the number of "good" outcomes and divide it by the total number of possible outcomes.

First, let's count the number of "good" outcomes, i.e., the number of ways we can draw three cards without any two of them differing by 1.

1. Start by choosing the first card. We have 10 options to choose from.
2. Then, for the second card, there are 7 cards left that do not differ by 1 from the first card (2, 3, 4, 6, 7, 8, 9).
3. Finally, for the third card, there are 6 cards left that do not differ by 1 from the second card chosen in step 2.

Thus, there are 10 * 7 * 6 = 420 "good" outcomes.

Now, let's count the total number of possible outcomes. We choose 3 cards from a deck of 10 without replacement, which can be done in 10C3 ways:

10C3 = (10 * 9 * 8) / (3 * 2 * 1) = 120.

Therefore, there are 120 possible outcomes.

Finally, we can calculate the probability by dividing the number of "good" outcomes by the total number of possible outcomes:

Probability = "good" outcomes / total outcomes = 420 / 120 = 7/2.

Hence, the probability that no two cards drawn have numbers that differ by 1 is 7/2 or 3.5.