A probability problem involving two cards (Part 2)

Here is a standard problem that could appear in an elementary probability class.

In yesterday’s post, I considered two different ways of solving a similar-looking problem, except the final word jack was replaced by ace. Yesterday, I showed that there were two legitimate ways of solving that problem, resulting (of course) in the same answer.

About a year ago, a student came into my office using these two different techniques to solve the ace/jack problem. Except she arrived at two different answers!

Method #1: One law for probability states that

Another law of probability states that

Combining these, we find that

Written more colloquially,

P(first an ace or second a jack)

= P(first an ace) + P(second a jack) – P(first an ace AND second a jack)

=P(first an ace) + P(second a jack) – P(first an ace) P(second a jack, given first an ace)

Let’s look at these three probabilities on the last line separately.

1. P(first an ace) is .
2. P(second a jack) is also . No information about the first card appears between the two parentheses, and so this is similar to pulling a card out of the middle of a deck. Since no information is given about the preceding card(s), the answer is still .
3. P(second an a jack, given first an ace) is different than the answer to #2 above. For this problem, the first card is known to be an ace, and the question is, given this knowledge, what is the probability that the second card is a jack? Since the first card is known to be an ace, there are still 4 jacks left out of 51 possible cards. Therefore, the answer is .

Putting these together, we find the final solution of

Here’s was the student’s second solution.

Method #2:There are three ways that either the first or second card could be an a jack:

1. The first card is an ace and the second card is a jack.
2. The first card is an ace and the second card is not a jack.
3. The first card is not an ace and the second card is a jack.

Each of these can be computed using the rule in much the same way as above:

1. P(first an ace) P(second a jack, given first an ace)
2. P(first an ace) P(second not a jack, given first an ace)
3. P(first not an ace) P(second a jack, given first not an ace)

Adding these together, we obtain the answer:

So, my student asked me, “Which one is the right answer? And why is the wrong answer wrong?” I must admit that it took me a couple of minutes before I found the student’s mistake.After all, the student’s logic perfectly paralleled the correct logic given in yesterday’s post.

I’ll discuss the mistake in tomorrow’s post. Until then, here’s a green thought cloud so that you also can think about what the student did wrong.