11 Plus Exams Forum

Hopefully someone can help me out with this, thanks.
Sineads pencil case has 2 pens and 5 pencils. Joans pencil case has 6 pens and 3 pencils. An item is taken from each pencil case in turn.
Draw a tree diagram to find all lossible outcomes.
What us the probability of getting: a pen follower by a pencil, and at least one pencil.
Thanks!

Well, I don't know what a probability tree is supposed to look like, but I'd have expected from the root you'd have a 'split' (two branches) to represent the first outcome (selecting an item from Sinead's pencil case) and then at the end of each branch you'd have another 'split' (two branches) to represent the second outcome (selecting an item from Joan's pencil case). On each branch you'd write the likelyhood of that event. I'm not sure how easily I can draw that here, but here goes:
In probability, an "and" is handled via multiplication. So the end of each 'route' through tree (i.e. at the bottom) can then be labelled with the result of multiplying the two fractions together. So reading left-to-right above, the four outcomes would be 2/7 x 6/9, then 2/7 x 3/9, then 5/7 x 6/9 and finally 5/7 x 3/9.

So the "a pen follower by a pencil" can be read from the result at the bottom of that 'route' and then "at least one pencil" answer can be established by adding the results at the bottom of each 'route' which contains a pencil.

I hope I've explained that sufficiently clearly.

Tree diagram start from a common point and branch horizontally.

Each outcome is calculated on the relevant 'branch'.

At least on pencil is more easily found by calculating 'no pencil' and then subtracting from 1 ... quicker than adding multiple fractions.

Never simplify fractions on a tree diagram as it makes any subsequent addition harder.

Is this an 11+ question?