A bag contains 7 green counters and 3 purple counters.
A counter is taken at random and its colour noted. The counter is not returned to the box.
Then a second counter is taken at random and its colour noted. It also is not returned to the box.
Finally a third counter is taken at random, and its colour noted.
Work out the probability that
all three counters are purple
exactly one of the three counters is purple
at least one of the three counters is purple
This is an "AND" question: 1st purple AND 2nd purple AND 3rd purple.
Each time a purple is chosen there is one less purple left, so the numerator goes down by one each time.
And each time a counter is chosen the total number remaining in the bag is one less, so the denominator also goes down by one each time.
This is an "AND" and "OR" question: [ 1st purple AND 2nd green AND 3rd green ] OR [ 1st green AND 2nd purple AND 3rd green ] OR [ 1st green AND 2nd green AND 3rd purple ].
The denominator needs to go down by 1 each time.
And the numerators need to change each time based on how many of each colour are left after previous counters have been taken.
It would be quicker to do this using the 'shortcut method'.
There are 3 ways to get exactly 1 purple counter (PGG, GPG, and GGP).
So the probability is .
The easiest way to do this is to realise that 'at least 1 purple' is the same as 'NOT all green'.
So find the 'all 3 green' probability and subtract it from 1.
'All three green' is an "AND" question: 1st green AND 2nd green AND 3rd green.
The numerators and denominators will change just as in part (i).
A large box contains 30 bags of crisps. There are 9 bags of ready salted crisps (R) , 17 bags of salt and vinegar crisps (V), and 4 bags of cheese and onion crisps (C) .
Aram takes at random two bags of crisps from the box.
Work out the probability that the two bags he takes are of different types.
There are several ways to do this.
The long way is as an AND/OR question considering all the different options: '[R AND V] OR [V AND R] OR [R AND C] OR [C AND R] OR [V AND C] OR [C AND V]'.
Be sure to change the numerators and denominators in the fractions to match what is left after each choice.
You could also use the 'shortcut method' to simplify this slightly:
'2
[R AND V] + 2
[R AND C] + 2
[V AND C]'.
The simplest way, however, is to do this as an AND/OR question with the following breakdown:
'[R AND not R] OR [V AND not V] OR [C AND not C]'
For [R AND not R] there is 9/30 probability of the first bag being R. Then there are 17+4=21 V and C bags left in the box, and 29 bags in total left in the box. So there is a 21/29 probability of the second bag NOT being R.
Do the same thing to find the numerators and denominators for [V AND not V] and [C AND not C].
Note that for each of the products, the numerators add up to 30 (the total number of crisps in the box to start).