Drawing with and without putting back

A marble is drawn from a bag, its colour is seen, and then another marble is drawn. This can be done two ways:

• With putting back: the first marble goes back into the bag before the second draw. The bag is as it was at the start, and the second draw is like the first.
• Without putting back: the first marble stays out. The bag holds one marble less, and what is left in it depends on the marble drawn.

The bag in the examples:
A bag holds three red marbles and two blue ones.
Two marbles are drawn from it, one after the other.

With putting back, the branches of the second draw are the same as those of the first:

• In each draw, the probability of red is $\displaystyle \frac{3}{5}$, and of blue $\displaystyle \frac{2}{5}$.
• The probability of two red marbles: $\displaystyle {\frac{3}{5}\cdot\frac{3}{5}}={\frac{9}{25}}$.

Without putting back, the probabilities in the second draw depend on the marble drawn first. The example below builds the tree diagram a stage at a time:

• The first draw is as with putting back: $\displaystyle \frac{3}{5}$ red and $\displaystyle \frac{2}{5}$ blue.
• After a red, the bag holds four marbles: two red and two blue. So $\displaystyle \frac{2}{4}$ and $\displaystyle \frac{2}{4}$.
• After a blue, three red and one blue are left: $\displaystyle \frac{3}{4}$ red and $\displaystyle \frac{1}{4}$ blue.
• A path’s probability is still the product of its branches. For two red marbles: $\displaystyle {\frac{3}{5}\cdot\frac{2}{4}}={\frac{6}{20}}$ – in $\displaystyle \frac{3}{5}$ of the times the first marble is red, and in $\displaystyle \frac{2}{4}$ of those the second is red too.

Without putting back, two marbles of the same colour are less likely: after a marble of some colour, the bag holds fewer of that colour. The example below finds, both ways, the probability that the two marbles are the same colour – both red or both blue. The tree shows the event’s paths without putting back:

• With putting back: $\displaystyle {\frac{3}{5}\cdot\frac{3}{5}+\frac{2}{5}\cdot\frac{2}{5}}={\frac{9}{25}+\frac{4}{25}}={\frac{13}{25}}={0.52}$
• Without putting back: $\displaystyle {\frac{3}{5}\cdot\frac{2}{4}+\frac{2}{5}\cdot\frac{1}{4}}={\frac{6}{20}+\frac{2}{20}}={\frac{8}{20}}={0.4}$

Drawing two pupils from a class is always without putting back: the same pupil cannot come out twice. So the second draw has one pupil less, and its probabilities depend on who came out first.

A bag holds two red marbles and three yellow ones.
A marble is drawn, not put back, and another marble is drawn.
Write the probabilities on the branches and each path’s probability.
Mark the path of the event “both marbles are yellow”, and find its probability.

A bag holds two red marbles and three yellow ones.
A marble is drawn, put back into the bag, and another marble is drawn.
Write the probabilities on the branches and each path’s probability.
Mark the paths of the event “the marbles are different colours”, and find its probability.

A bag holds one green marble and three white ones.
Two marbles are drawn from it, one after the other.
Find the probability that both marbles are white:

With putting back.

Without putting back.

What is the probability that both marbles are green – with putting back, and without? Explain.

The class committee has four girls and three boys.
Two pupils are drawn from the committee to represent the class on the pupils’ council.
Write the probabilities on the branches and each path’s probability.
Mark the paths of the event “a girl and a boy are drawn”, and find its probability.

A bag holds four red marbles and two blue ones.
Two marbles are drawn from it without putting back.
Find the probability that at least one of them is blue.