Independent events

Two events are independent when knowing that one of them happened does not change the probability of the other. When it does change it, the events are dependent.

A fair die is thrown once.
$\displaystyle A$ – an even number comes up: $\displaystyle \{2, 4, 6\}$, so $\displaystyle {P(A)}={\frac{3}{6}}={\frac{1}{2}}$.

• $\displaystyle B$ – a number greater than 3 comes up: $\displaystyle \{4, 5, 6\}$. If $\displaystyle B$ is known to have happened, three outcomes are left, and two of them are even: $\displaystyle {P(A|B)}={\frac{2}{3}}$. The probability changed, so $\displaystyle A$ and $\displaystyle B$ are dependent.
• $\displaystyle C$ – a number greater than 4 comes up: $\displaystyle \{5, 6\}$. If $\displaystyle C$ is known to have happened, two outcomes are left, and one of them is even: $\displaystyle {P(A|C)}={\frac{1}{2}}$. The probability did not change, so $\displaystyle A$ and $\displaystyle C$ are independent.

Each of these three equalities says that $\displaystyle A$ and $\displaystyle B$ are independent, and checking one of them is enough:

• The first: knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$.
• The second: the probability of “both $\displaystyle A$ and $\displaystyle B$” is the product of the probabilities.
• The third: $\displaystyle A$ has the same probability when $\displaystyle B$ happened and when $\displaystyle B$ did not.

On the die, $\displaystyle A$ – even and $\displaystyle C$ – greater than 4:

• $\displaystyle {P(A\cap C)}={P(\{6\})}={\frac{1}{6}}$, and also $\displaystyle {P(A)\cdot P(C)}={\frac{1}{2}\cdot\frac{1}{3}}={\frac{1}{6}}$.
• $\displaystyle {\overline{C}}={\{1, 2, 3, 4\}}$, and two of its outcomes are even: $\displaystyle {P(A|\overline{C})}={\frac{2}{4}}={\frac{1}{2}}={P(A|C)}$.

The proof below shows that each of the three equalities follows from the others, so there is no need to check all three. We assume that $\displaystyle 0<P(B)<1$.

The first and the second. By the definition of conditional probability, $\displaystyle {P(A|B)}={\frac{P(A\cap B)}{P(B)}}$. Multiplying by $\displaystyle P(B)$, we get that $\displaystyle {P(A|B)}={P(A)}$ exactly when $\displaystyle {P(A\cap B)}={P(A)\cdot P(B)}$.

The first and the third. By the law of total probability:

• $\displaystyle P(A)$ is a weighted mean of $\displaystyle P(A|B)$ and $\displaystyle P(A|\overline{B})$, since $\displaystyle {P(B)+P(\overline{B})}={1}$.
• If the two conditional probabilities are equal, their weighted mean equals them too: $\displaystyle {P(A)}={P(A|B)}$.
• If $\displaystyle {P(A|B)}={P(A)}$, substituting in the formula gives $\displaystyle {P(\overline{B})\cdot P(A|\overline{B})}={P(A)-P(B)\cdot P(A)}={P(\overline{B})\cdot P(A)}$. Dividing by $\displaystyle P(\overline{B})$: $\displaystyle {P(A|\overline{B})}={P(A)}={P(A|B)}$.

Disjoint events, each with a positive probability, are always dependent: if one of them happened, the other can no longer happen.

On the die: $\displaystyle A$ – even, $\displaystyle D$ – odd.

• $\displaystyle {A\cap D}={\emptyset}$, so $\displaystyle {P(A\cap D)}={0}$.
• $\displaystyle {P(A)\cdot P(D)}={\frac{1}{2}\cdot\frac{1}{2}}={\frac{1}{4}}$, and $\displaystyle 0\neq\frac{1}{4}$.
• In words: if an odd number is known to have come up, the probability of an even one fell from $\displaystyle \frac{1}{2}$ to 0.

“Disjoint” says the events do not happen together; “independent” says one does not affect the chance of the other. These are two different things.

When the same experiment is repeated, and each time’s result does not affect the other times, the stages are independent: every branch has the same probability whatever the path before it, and a path is the product of the probabilities.

A fair die is thrown three times.
$\displaystyle Y$ – an even number greater than 3 comes up, that is 4 or 6: in each throw $\displaystyle {P(Y)}={\frac{2}{6}}={\frac{1}{3}}$.
The tree marks the paths of “$\displaystyle Y$ in exactly two throws”:

• Three paths: $\displaystyle YYN$, $\displaystyle YNY$, $\displaystyle NYY$ – the throw without $\displaystyle Y$ can be the first, the second or the third.
• Each of them has the same probability: $\displaystyle {\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}}={\frac{2}{27}}$.

A year group has 200 students: who does sport, and who wears glasses.

A student is chosen at random. What is the probability that the student wears glasses?

What is the probability that a student wears glasses, given that the student does sport?

Are the events “wears glasses” and “does sport” independent? Explain using the answers.

$\displaystyle A$ and $\displaystyle B$ are independent events.
$\displaystyle {P(A)}={0.4}$, $\displaystyle {P(B)}={0.5}$.

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

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

$\displaystyle {P(A)}={0.3}$, $\displaystyle {P(B)}={0.6}$, $\displaystyle {P(A\cup B)}={0.72}$.

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

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

A fair die is thrown three times.
$\displaystyle Y$ – an even number greater than 3 comes up.

Write the probabilities on the branches and each path’s probability.

What is the probability that $\displaystyle Y$ comes up only in the first throw and the third?

What is the probability that $\displaystyle Y$ comes up in the first throw and the third (in the second – any result)?

A test has two exercises in probability and one in statistics. In each exercise the candidate passes or fails.
To pass the whole test, a candidate must pass at least two of the three exercises.
The probability of passing each probability exercise is $\displaystyle 0.6$.
The probability of passing the statistics exercise is $\displaystyle 0.8$.
The results of the different exercises are independent of each other.

In the tree, the first and second stages are the two probability exercises, and the third stage is the statistics exercise. Write the probabilities on the branches and each path’s probability.

What is the probability that a candidate passes all three exercises?

What is the probability that a candidate passes exactly two exercises and fails one?

What is the probability that a candidate passes the whole test?

A candidate passed the whole test. What is the probability that the candidate passed both probability exercises? Round to two decimal places.

$\displaystyle A$ and $\displaystyle B$ are disjoint events.
$\displaystyle {P(A)}={0.3}$, $\displaystyle {P(B)}={0.5}$.

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