Probability of Combined Events

A couple have two children. What is the probability that both are boys?

Solution

Given that there is an equal probability of having either a boy or a girl, the probability of one child being a boy is $P("Boy")=0.5$.

Since the outcome of the first child (whether the first child is a boy or a girl) doesn't effect the outcome of the second, we can say that the sex of each child is independent. Therefore:

$P("Twoboys")=P("Boy")\times P("Boy")=0.5\times 0.5=0.25$

You roll a standard 6-sided dice. What is the probability that you roll either a 3 or a 5?

Solution

Since you cannot roll both 3 and 5 at the same time, these two events are mutually exclusive. This means that we can apply the 'sum' rule:

$P("rolling3"\cup "rolling5")=P("rolling3")+P("rolling5")=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$

There are 4 identical red apples. Two of them are poisonous and the remaining two are harmless. The poison isn't very strong, however. You will only die if you eat both poisonous apples consecutively.

You eat three of the apples. What is the probability that you live?

Solution

Let us denote the poisonous apples as 'P' and the regular apples as 'A'. The combination of all possible outcomes is as follows:

AAP, APA, PAA, PAP, PPA, APP

You will die only if you consume two poisonous apples consecutively. Therefore, you will survive if the outcome is AAP, APA, PAA or PAP. Using the basic probability formula:

$P("Surviving")=\frac{4}{6}=\frac{2}{3}$

Therefore the probability of surviving is $\frac{2}{3}$

45% of pet owners own dogs, and 4% own both cats and dogs. Given that someone owns a dog, how likely is it that they also own a cat?

Solution

First, use the conditional probability formula to express the question mathematically:

Next, input the numbers from the question and compute the answer:

$P("Cat"|"Dog")=\frac{P("Cat"\cap "Dog")}{P("Dog")}=\frac{0.04}{0.45}=\frac{4}{45}=0.0889$ to 3 s.f.

The rate of infection of a particular disease in a population is 1%. The test for this disease is accurate 99% of the time. Demonstrate this information using a probability tree.

Solution

If the test is 99% accurate, this means that:

• if you test positive, there is a 99% chance you have the disease
• if you test negative, there is a 99% chance you don't have the disease

We can illustrate this as follows:

Tree diagram 1

The second column of branches, labelled "infected" and "healthy", indicate conditional probabilities.

For example, the outer branches indicate the probability that someone is infected given that they test positive:

Tree diagram 2

We can use this diagram to, for example, find the probability that someone is infected given that they test negative. Read off the diagram as follows:

Tree diagram 3

We can see that $P\left(Infected\right|Negative)=0.01$.

Another interpretation is that the information to the left of the "given that" symbol indicates the outcome of the first branch. The information to the right of the "given that" symbol indicates the outcome of the second set of branches.

Take the scenario in the example above. What is the probability that someone is both infected and tests negative?

Solution

First, reformulate the question using the conditional probability formula:

From our tree diagram above, we know that the probability that someone is infected given that they test negative is 0.01. We also know that the overall probability of someone testing negative is 0.99.

Input these probabilities and rearrange:

$P\left(Infected\right|Negative)=0.01=\frac{P(Infected\cap Negative)}{0.99}$

Therefore the probability of testing negative and being infected with the disease is 0.0099.