Probability trees, with and without replacement

What you will be able to do

• draw a probability tree and use it: multiply along a path, add across paths;
• find chances for draws with and without replacement;
• explain why draws from a big population can be treated as independent.

Two phones from a batch

A made-up batch holds $10$ phones, and $3$ of them are faulty. An inspector picks $2$ phones, one after the other, and does not put the first back. What is the chance that both are faulty?

The first phone is faulty with probability $\dfrac{3}{10}$. If it is, the batch now holds $2$ faulty phones among $9$. So the second is faulty with probability $\dfrac{2}{9}$. Both are faulty with probability

Probability trees

A probability tree draws each step as a set of branches.

• The branches from one point add up to $1$.
• Multiply along a path to get the chance of that whole path.
• Add the paths that make the event you want.

For the phones, the three results are both faulty $\dfrac{1}{15}$, exactly one faulty $\dfrac{7}{15}$ and none faulty $\dfrac{7}{15}$. Exactly one has two paths, faulty then good and good then faulty. Each is $\dfrac{3}{10} \times \dfrac{7}{9} = \dfrac{7}{30}$, and together they make $\dfrac{14}{30} = \dfrac{7}{15}$. Check: $\dfrac{1}{15} + \dfrac{7}{15} + \dfrac{7}{15} = 1$.

With and without replacement

If the inspector put the first phone back, the batch would be the same for the second pick. The picks would be independent, and both faulty would have probability $0.3 \times 0.3 = 0.09$. Without replacement it is $0.0667$. In a small batch, the difference is large.

Now take a big batch: $10000$ phones, with the same share of $30\%$ faulty. Without replacement,

That is almost exactly $0.09$. Taking one phone out of $10000$ hardly changes the share that is left. This is why a survey of $1000$ people from a whole country can treat the people as independent picks, even though nobody is asked twice. A common rule of thumb says this is safe when the sample is less than about a tenth of the population.

Try it in Python

The program works out the two phone answers in exact fractions, and then the big batch.

A tree with data on its branches

Trees are not only for bags and batches. A made-up record says that a train is late on $20\%$ of days. On a late day, a student misses the start of the first lesson with probability $0.5$. On other days, the chance is $0.05$. The two paths that end in "misses" give $0.2 \times 0.5 + 0.8 \times 0.05 = 0.1 + 0.04 = 0.14$.

Your turn

The usual mistakes

• Keeping the first chance for the second draw. Without replacement, both the top and the bottom of the fraction change.
• Adding along a path. Along a path you multiply. You add only different paths.
• Missing an order. "Exactly one faulty" can happen as faulty then good, or good then faulty.
• Worrying about replacement in a huge population. Then the two kinds of draw give almost the same answer.

Remember these

• In a tree, the branches from a point add up to $1$; multiply along a path; add the paths.
• Without replacement, the branches after the first draw use the new counts.
• From a big population, draws without replacement are almost independent.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A bag holds $4$ red and $6$ white beads. Two are taken without replacement. Find the probability that both are white.

2. The same bag is used, but the first bead is put back. Find the probability that both are white.

3. In a tree, the branches from one point are $0.35$, $0.4$ and a third branch. What is the third branch?

4. A made-up company has $8000$ workers, and $20\%$ of them cycle to work. Two workers are picked at random without replacement. About how likely is it that both cycle?

5. Why do the paths of a tree always add up to $1$?

References

• OpenStax, Introductory Statistics, Chapter 3, the sections on independent events and on tree and Venn diagrams.