I gave the question below out as homework to 2 AS level students (Edexcel AS and A level Modular Mathematics: S1, p84):

'On a firing range, a rifleman has two attempts to hit a target. The probability of hitting the target with the first shot is 0.2 and the probability of hitting with the second shot is 0.3. The probability of hitting the target with both shots is 0.1.
Find the probability of:
a) missing the target with both shots,
b) hitting the target with the first shot and missing with the second.

One student used Venn diagrams and got:
a) 0.6
b) 0.1

The second student used tree diagrams and got:
a) 0.56
b)0.14

Bearing in mind that statistical questions often ask for 2/3 decimal places in their answers, how would you have explained this discrepancy?

I can see that, using the tree diagram, the probability of hitting the target with both shots becomes 0.06, not 0.1 as given. However we are given 'the probability of hitting with the second shot is 0.3.' Presumably the second set of branches on the tree diagram can't be labelled with 0.3 and 0.7 but I'm not sure how you justify changing 0.3 to 0.25 or 0.7 to 0.75, which presumably you had to in order to get the answer you've given. But even that isn't correct since 0.2 x 0.25 = 0.05, which still isn't the 0.1 given.

I presume the tree diagram must be set up in a slightly different way to the simple idea of first shot then branches for second shot? As you should be able to use a tree diagram to calculate this answer I'm wondering how you split the two sets of branches up?

I've never bothered using tree diagrams for A level questions before, I always use Venn diagrams, so I'm a little bit bothered that a seemingly simple question is proving so difficult.

The tree diagram is as normal. First the first shot, hit or miss, and then the second shot , hit or miss, giving 4 branches altogether,

The two branches which lead to hitting with the second shot must add up to 0.3.

And the branch which leads to hitting both targets must equal 0.1

So we have 0.2P(S/F) = 0.1 for the hitting both targets, P(S/F) = 0.5 , were P(S/F) is the probability of hitting with the second shot, provided the first has hit.

P(F') = 0.8

And we have 0.8P(S/F') + 0.1 = 0.3, so the probability to go on that branch for the second shot is 0.25.

So the chance of both missing will be 0.8 x 0.75 = 0.6

I have since looked at questions using both tree diagrams and Venn diagrams, for illustration purposes, and they all involve this kind of arguement. The only clue, that I can see, that this kind of care needs to be taken is that 'The probability of hitting the target with both shots is 0.1.' Clearly this doesn't happen if you take the simple GCSE tree diagram stand point.

I'm going to work on a few problems, using both methods until I've got this sorted out in my own mind. Then I'll try working through this with my students. The simple answer may be to tell them to only use Venn diagrams!