Problem:
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.
Scott's solution:
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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
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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.
To find the probability that no two of the cards drawn have numbers that differ by 1, we can use a different approach:
First, let's find the total number of ways to choose 3 cards out of 10 without replacement. This is given by the combination formula, written as 10C3, which is equal to 10! / (3!(10-3)!) = 120.
Next, let's find the number of favorable outcomes, which are the ways of choosing 3 cards without any pair differing by 1. To do this, we can consider the 6 pairs of consecutive numbers: (1,2), (2,3), (3,4), (4,5), (5,6), and (6,7). For each pair, we have 8 remaining cards to choose for the third card (excluding the two cards in the pair). So, the number of favorable outcomes is equal to 6 * 8 = 48.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes. Thus, the probability is 48 / 120 = 4/10 or 2/5.
It appears that the answer key you reviewed may be incorrect, as the correct probability is 2/5 and not 7/15.