<p>Six index cards are marked so that each has a different integer from 1 to 6, inclusive, on it. If two different cards are chosen at random, what is the probability that the number on the second card is one greater than the number on the first card?</p>
<p>The answer is 1/6 but I don't know how that number was reached.</p>
<p>There are 5 possible combinations of cards (1,2),(2,3),(3,4),(4,5),(5,6) and the probability for all of them is the same i.e. 1/6<em>1/5. This is because you take the first card out when you choose it leaving only 5 behind. So 5</em>1/30 is 1/6</p>
<p>To compute a simple probability where all outcomes are equally likely, divide the number of “successes” by the total number of outcomes.</p>
<p>The total number of outcomes is 6*5=30 (I used the counting principle here).</p>
<p>5 of these outcomes are successes - (1,2), (2,3), (3,4), (4,5), (5,6).</p>
<p>So the probability is 5/30 = 1/6.</p>
<p>Remarks:</p>
<p>(1) There are at least 3 ways to compute the total: </p>
<p>(a) by listing all the possibilities - </p>
<p>(1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,3), (2,4),…
…</p>
<p>(b) Using the counting principle as I did above (the counting principle says that if one event is followed by a second independent event, the number of possibilities is multiplied).</p>
<p>(c) Using a permutation - in this case the total is 6P2. This computation can be done right on your calculator. We use a permutation here, and NOT a combination, because if you choose two cards in a different order you get a different possibility - for example (1,2) and (2,1) are different possibilities (in fact, (1,2) is a “success” whereas (2,1) is a “failure”).</p>
<p>Thanks! Both your explanations are clearer than the one my teacher gave me.</p>