Dependent and independent events

Two events are independent when knowing that one of them happened adds no information about the other – does not change its probability:

Otherwise the events are dependent. The definition is symmetric: if $\displaystyle {P(A/B)}={P(A)}$, then also $\displaystyle {P(B/A)}={P(B)}$.

A fair die is thrown. $\displaystyle A$ – “an even number comes up”, $\displaystyle {P(A)}={\frac{1}{2}}$.

• $\displaystyle B$ – “at most 2 comes up”, $\displaystyle {B}={\{1,2\}}$: of the two outcomes of $\displaystyle B$ one is even, $\displaystyle {P(A/B)}={\frac{1}{2}}={P(A)}$. $\displaystyle A$ and $\displaystyle B$ are independent.
• $\displaystyle C$ – “at least 4 comes up”, $\displaystyle {C}={\{4,5,6\}}$: of the three outcomes of $\displaystyle C$ two are even, $\displaystyle {P(A/C)}={\frac{2}{3}\neq P(A)}$. $\displaystyle A$ and $\displaystyle C$ are dependent.

Independence can be tested in three ways, each equivalent to the others: if one holds, they all hold.

• $\displaystyle {P(A/B)}={P(A)}$ – the definition.
• $\displaystyle {P(A/B)}={P(A/\overline{B})}$ – the probability of $\displaystyle A$ is the same when $\displaystyle B$ happens and when it does not.
• $\displaystyle {P(A\cap B)}={P(A)\cdot P(B)}$ – the handiest test: no division, and it holds even when $\displaystyle {P(B)}={0}$.

The third test follows from the definition of conditional probability: $\displaystyle {P(A\cap B)}={P(B)\cdot P(A/B)}$, and if $\displaystyle {P(A/B)}={P(A)}$, then $\displaystyle {P(A\cap B)}={P(B)\cdot P(A)}$.

With the die: $\displaystyle {P(A\cap B)}={P(\{2\})}={\frac{1}{6}}={\frac{1}{2}\cdot\frac{1}{3}}$ – independent; $\displaystyle {P(A\cap C)}={P(\{4,6\})}={\frac{1}{3}\neq\frac{1}{2}\cdot\frac{1}{2}}$ – dependent.

In a table of numbers, two events are independent when a column’s share is the same in every row – and the same as its share of the total.

200 students were asked whether they like maths:

• Among the boys: $\displaystyle {\frac{48}{120}}={0.4}$; among the girls: $\displaystyle {\frac{32}{80}}={0.4}$; among all the students: $\displaystyle {\frac{80}{200}}={0.4}$.
• “The student likes maths” and “the student is a boy” are independent: knowing that the student is a boy does not change the chance of liking maths.

Disjoint events and independent events are two different things. Two disjoint events, each with a probability greater than 0, are always dependent:

• they are disjoint, so $\displaystyle {P(A\cap B)}={0}$;
• $\displaystyle P(A)>0$ and $\displaystyle P(B)>0$, so $\displaystyle P(A)\cdot P(B)>0$;
• $\displaystyle P(A\cap B)\neq P(A)\cdot P(B)$.

In words: if $\displaystyle B$ is known to have happened, then $\displaystyle A$ is known for certain not to have happened – $\displaystyle {P(A/B)}={0}$. That is a great deal of information about $\displaystyle A$.

When events are known to be independent, the probability of the intersection is a product. The complements of independent events are independent too, so “at least one” is worked out by the complement.

A factory has two machines that work independently of each other.
The probability that machine A breaks down on a given day is 0.1, and the probability that machine B breaks down is 0.05.

• Both machines break down: $\displaystyle {0.1\cdot0.05}={0.005}$.
• At least one machine breaks down – the complement of “no machine breaks down”: $\displaystyle {1-0.9\cdot0.95}={1-0.855}={0.145}$.

A fair die is thrown.
$\displaystyle A$ – “a number greater than 4 comes up”; $\displaystyle B$ – “an odd number comes up”; $\displaystyle C$ – “a prime number comes up” (2, 3 or 5).

Work out $\displaystyle P(A\cap B)$.

Are $\displaystyle A$ and $\displaystyle B$ independent? Explain with $\displaystyle {P(A\cap B)}={P(A)\cdot P(B)}$.

Are $\displaystyle B$ and $\displaystyle C$ independent? Explain.

In a survey of 500 drivers:

A driver is known to have used a phone while driving. What is the probability that the driver had an accident?

Are “the driver had an accident” and “the driver used a phone” independent events? Explain.

$\displaystyle A$ and $\displaystyle B$ are independent events.
$\displaystyle {P(A)}={0.4}$, $\displaystyle {P(A\cup B)}={0.7}$.

Write $\displaystyle {P(B)}={x}$, and find $\displaystyle x$.

What is the probability that neither $\displaystyle A$ nor $\displaystyle B$ happens?

A house has three smoke detectors, each working independently of the others.
The probability that a given detector sounds the alarm in a fire is 0.9.

What is the probability that at least one detector sounds the alarm in a fire?

$\displaystyle A$ and $\displaystyle B$ are independent events, each with the probability $\displaystyle p$.
$\displaystyle {P(A\cup B)}={0.64}$.

Write an equation in $\displaystyle p$ and solve it. Only a solution between 0 and 1 can be a probability.

$\displaystyle {P(A)}={0.3}$, $\displaystyle {P(B)}={0.5}$.

Work out $\displaystyle P(A\cup B)$ if $\displaystyle A$ and $\displaystyle B$ are disjoint events.

Work out $\displaystyle P(A\cup B)$ if $\displaystyle A$ and $\displaystyle B$ are independent events.