Bradley's Maths

Solution

This is a fairly straightforward probability question that is likely to turn up in the middle of one of your exam papers and it can be either calculator or non-calculator. The only difficulty with this question is I haven't provided a diagram. In the exam you will usually get one that you need to fill in some of the probabilities on, but if they don't give you a tree diagram, sketch one for yourself - that is exactly what I did in the video, its quick and its simple.

The diagram is shown below:

The diagram doesn't need to be fabulous just a rough sketch like this is perfectly acceptable. There are no marks for it but it is enough to help you get full marks for the question.

Part a) asks for the probability of two reds. This is the top line in the tree and we multiply along the lines. We should remember that this is without replacement and so the second fractions have all been altered - there is one less sweet of whatever colour was taken and there is one less sweet in the bag.

$$P(r, r)= \dfrac{3}{8}\times\dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}$$

Part b) asks for the probability of one of each. There are two ways to take one of each sweet, first red then green or first green then red. Once again we multiply along the lines but we add between the lines

$$P(\text{one of each}) = \dfrac{3}{8}\times\dfrac{5}{7} + \dfrac{5}{8}\times\dfrac{3}{7} = \dfrac{30}{56} = \dfrac{15}{28}$$

Finally in part c) we are looking for the probability of at least one of the sweets being green. There are two ways to look at this problem, the hard way is to work out the probability of the first sweet being green, add the probability of the second being green and then add the probability of both sweets being green. The easier way to look at it is, if one has to be green they cannot both be red, so we take the probability of both red (which we worked out in part a) away from 1 (the sum of all probabilities).

$$P(\text{at least one green})= 1 - P(r,r) = 1 - \dfrac{3}{28} = \dfrac{25}{28}$$