Replacement and non-replacement probabilities

Consider the following experiments:

• drawing two cards
• picking three marbles from a bag
• drawing names from a hat

What these experiments have in common is that they involve multiple trials. After the first card has been drawn, the first marble has been picked, or the first name has been drawn, there are two possibilities before carrying out the second trial.

• Return the card/ball/number to the bag to be considered in the second draw (REPLACE the object)
• Keep the card/ball/number out and draw from the remaining options (DO NOT REPLACE the object)

More formally, we call each of these cases WITH REPLACEMENT or WITHOUT REPLACEMENT.  

Whether we return the item or keep it out affects the resulting probabilities.

Let's consider a simple example.

Example

A bag of marbles contains $3$3 red and $3$3 blue marbles. $2$2 marbles are drawn one after the other.

Calculate the probability of getting $2$2 reds:

a) if we replace the first before drawing the second (with replacement)

b) without replacing the first marble selected (non-replacement)

Part a: with replacement

When we draw the first marble, we have a $\frac{3}{6}$36​ chance of getting a red marble.

If we return this red marble to the bag, then everything is reset. The probability that we pick a red marble on the second go will also be  $\frac{3}{6}$36​.

Multiplying these two together, we get that P(Red, Red) = $\frac{3}{6}\times\frac{3}{6}=\frac{1}{4}$36​×36​=14​ = $25%$25%

Part b: without replacement

Like before, when we draw the first marble, we have a $\frac{3}{6}$36​ chance of getting a red marble.

Now look what happens if we keep this ball out:

• There will only be $5$5 marbles left to choose from.
• Since we are considering the event of drawing two red marbles, we assume that the first one we picked was red, so there will only be $2$2 red ones to choose from on the second go. 
• This means that on the second draw, we have a $\frac{2}{5}$25​ chance of getting a red marble. 

Multiplying these two together, we get that P(Red, Red) = $\frac{3}{6}\times\frac{2}{5}=\frac{1}{5}$36​×25​=15​ = $20%$20%

So whether we do or do not replace the object before the next draw affects the overall probability of the event.

Examples

Question 1

A container holds four counters coloured red, blue, green and yellow.

a) Draw a tree diagram representing all possible outcomes when two counters are drawn, and the first counter is replaced before the next draw.

b) Draw a tree diagram representing all possible outcomes when two counters are drawn, and the first counter is not replaced before the next draw.

Question 2

A number game uses a basket with 9 balls, all labelled with numbers from 1 to 9. Three balls are drawn at random without replacement.

a) What is the probability that the ball labelled '3' is picked once?

b) What is the probability that the ball labelled '3' is picked and the ball labelled '6' is also picked?