The letters A,B,C,D,E,F, and G are arranged in random order.

Question

The letters A,B,C,D,E,F, and G are arranged in random order.

(a) What is the probability that A and E are next to each other?

(b) What is the probability that A, B, and C appearing in a row in some

order in the string of letters?

Answer 1

Julie,

 

The easiest ay to think about probability is if you can specify all the possible equally likely outcomes and then pick out those that satisfy your question, then

 

Desired probability = [Sum of picked outcomes] / [Total of possible outcomes].

 

In your problem, there are 7 letters (ABCDEFG) that can be arranged in random order.  No letter can appear more than once in the 7-letter sequence.  Thus, there are 7 possibilities for a letter in position1, 6 remain for position2, then 5 remain for position3, etc.  Then the total number of the se 7-letter sequences possible is 7 factorial (7!) which is just 7*6*5*4*3*2*1 = 5,040.  That's your denominator  Each of these sequences is equally likely because of the specified randomness.  Now to handle your 2 questions:

 

(a) What's the probability that A & E are next to each other?

 

Let's sketch out the possibilities:

1. Suppose A is in position1, then E must be in position2 and we don't care how the remaining 5 letters are arranged (that's 5!  or  120 possibilities of the 5040).

2. Now suppose A appears in position 2, then E has to be in either position1 or position3, and again we don't care about the other 5 letters, so that's 2 sequences of 120 possibilities each, or 240 A in position2.

3. Repeating this argument for A in positions3 thru 6 and E on either side of A.  That's 4 cases of 240 each (as in #2) or 960 possibilities that satisfy our question.

4. Finally, when A is in position7, E can only be in position6, so that gives us 120 cases as in #1.

 

Therefore, the Probability for question (a) is = [120 + 240 + 960 + 120] / 5040 = 1440 / 5040 = 0.286

 

(b) What's the probability that A B & C are next to each other in some order?

Note that you didn't mean to use the word appear since they are guaranteed to be in every sequence (probability = 1) by definition of how the sequences are constructed.  So the important point is they have to be next to each other.

Therefore, we can use a similar logical approach as in part (a);

 

1. If A is in position1, then ABC or ACB satisfy our question, and we don't care about the order of the other 4 letters (4! or 24 possibilities).  So we have two cases of 24 each for a total of 48 here.

2. If A is in position2, then BAC, CAB, ABC & ACB satisfy our question as in giving us 96 more sequences.

3. If A is in position3, then BCA, CBA, BAC, CAB, ABC & ACB give us 6 * 24 or 144 sequences.

4. Continuing this logic by placing A in position4 thru 7 will produce (6 + 6 + 4 + 2)*24 = 432 more sequences. Notice the symmetry position 1 & 2 and then 6 & 7 because it gets near the extremes which cuts down possibilities.

 

Thus, the Probability of condition (b)  = [48 + 96 + 144 + 432] / 5040 = 720 / 5040 = 0.143

 

So by specifying a 3 letter condition versus a two-letter condition, we've reduced the number of 7-letter sequences drawn randomly that will satisfy the condition, so the probability must decrease.

 

That's a lot of work to keep the details straight, but the approach is straightforward in principle. Good luck!