Mathematics 43 Online
OpenStudy (anonymous):

the first 12 alphabets are written down at random. what is the probability that A and B are written down side by side a. 1/5 b. ¼ c. 1/6 d. 1/8

OpenStudy (anonymous):

I would do it like this. There are 2 different options: a, A or B is the first or last in the list b, A or B is somewhere in the middle. a, It is 2/12=1/6 chance that lets say A is the first or the last. Than there is only 1 way way that they can be next to each other. So the chance is 1/11 b, It is 10/12=5/6 chance that lets say A is not the first/last. Than there are two ways they can be next to each other (in front/behind). So the chance is 2/11. we have to add these. 1/6*1/11+5/6*2/11=11/66=1/6

OpenStudy (anonymous):

Do those account for B being first or last in part a (and conversely not first/last in b)?

OpenStudy (anonymous):

I dont get you, sorry.

OpenStudy (anonymous):

I just thought of placing them 1 by 1. So first I only cared about the position of A than counted the chance that B is next to it.

OpenStudy (anonymous):

Probability is my favourite maths subject! You have to think a lot, and it is just pure logic

OpenStudy (anonymous):

thank u :)

OpenStudy (anonymous):

In pasrt a) of your solution you mention that there is a 1/6 chance of A being in the first or last position. But it's possible that you have "BA__________" or "__________AB" which I'm not sure is covered in your case. I love probability too...but I've been out of it for enough years that I'm losing my edge. You might be right...I might just not be seeing it.

OpenStudy (anonymous):

Oh I get you, but I dont think that this matters here. (I am 98% confident :-))

OpenStudy (anonymous):

It might not...looking at it more, part B covers the cases where A is *not* first/last...which would include my example. You then place B on either side...which might include it being in the first or last position (depending on A being #2 or #11). Ok..I can conceed :) Good solution!

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