Combinations and probability

A netball team of 7 players is to be chosen from a list of 12.
(i) How many different teams of 7 can be chosen? [1]
(ii) The list of 12 includes Anna and Annette. A team of 7 people is chosen at random. Given that at least one of Anna and Annette is chosen, find the probability that both of them are chosen. [3]

I've done the first part.

But on part 2 I am getting 6/11 when the answer is 3/8.

My reasoning is.. if one of A1 or A2 is chosen then we can just forget about them for now. So.. the question would become..

I have 11 people. I need to choose 6. What's the probability one specific person, A, is chosen?

So I did 10C5/11C6 (total teams of 6 with A/ total teams) and got 6/11.

Where am I going wrong? I have a feeling it is something to do with the fact that the 'given' person could be A1 or A2, but I just can't get my head around it.

You're not accounting for the 'at least one' bit. You're assuming 'exactly one'.

How could I account for that?

Actually now that I've worked through the problem, I get 6/11 as well. So either I've made a mistake (very possible - did it on a piece of scrap quickly) somewhere or the book is incorrect.

I believe the book is correct. I think it's clear the difficulty is in evaluating the number of ways of choosing a group with at least one of Anna and Annette.

I did this as "# of groups with Anna" + "# of groups with Annette" - "#of groups with Anna + Annette".

(Note: I also did a quick check via computer and that also gives 3/8).

Actually now that I've worked through the problem, I get 6/11 as well. So either I've made a mistake (very possible - did it on a piece of scrap quickly) somewhere or the book is incorrect.

You want

$P(A_1 \cap A_2 | (A_1 \cap A_2) \cup (A_1 \cap \bar{A}_2) \cup (\bar{A}_1 \cap A_2)) \\ = \frac{n(A_1 \cap A_2)}{ n( A_1 \cap A_2) \cup (A_1 \cap \bar{A}_2) \cup (\bar{A}_1 \cap A_2))}$

where $A_1, A_2$ are Anna and Annette. This seems to work out to 3/8

You want

$P(A_1 \cap A_2 | (A_1 \cap A_2) \cup (A_1 \cap \bar{A}_2) \cup (\bar{A}_1 \cap A_2)) \\ = \frac{n(A_1 \cap A_2)}{ n( A_1 \cap A_2) \cup (A_1 \cap \bar{A}_2) \cup (\bar{A}_1 \cap A_2))}$

where $A_1, A_2$ are Anna and Annette. This seems to work out to 3/8

[edit: beaten by others, but latex of this quality shouldn't be deleted lightly]

Yep, my mistake was to treat as just one $A_1 \cap \hat{A}_2$ for whatever reason was originally going on through my head at the time...

Ta.

Going to be a git and point out your LaTeX would be improved by using \dfrac instead of \frac:

You can also get to 672 via 12C7 - 10C7, where 10C7 is the number of teams with neither Anna nor Anette.