What is the probability that someone with black hair has brown eyes?

\(\text{80}\) people have black hair and of those, \(\text{50}\) people also have brown eyes.
Therefore the probability that someone with black hair has brown eyes is \(\frac{50}{80} = \frac{5}{8}\).

**Note:** this is different from asking for the probability of having black hair **and** brown eyes.
(This probability is computed in part (d) below.)
The question was phrased to ask for the probability of having brown eyes **given** that a person has black hair.

What is the probability that someone has black hair?

Out of a total of \(\text{230}\), \(\text{80}\) have black hair.
Therefore the probability that someone has black hair is \(\frac{80}{230} = \frac{8}{23}\).

What is the probability that someone has brown eyes?

Out of a total of \(\text{230}\), \(\text{120}\) have brown eyes.
Therefore the probability that someone has brown eyes is \(\frac{120}{230} = \frac{12}{23}\).

Are having black hair and having brown eyes dependent or independent events?

We already computed that the probability of having

- black hair is \(\frac{8}{23}\); and
- brown eyes is \(\frac{12}{23}\).

Since \(\text{50}\) out of \(\text{230}\) people have black hair and brown eyes, the probability of having black hair and brown eyes is \(\frac{5}{23}\).

We conclude that having black hair and brown eyes are dependent events since \(\frac{5}{23} \ne \frac{8}{23} \times \frac{12}{23}\).

Given the following contingency table, identify the events and determine whether they are dependent or independent.

| Location A | Location B | Totals |

Buses left late | \(\text{15}\) | \(\text{40}\) | \(\text{55}\) |

Buses left on time | \(\text{25}\) | \(\text{20}\) | \(\text{45}\) |

Totals | \(\text{40}\) | \(\text{60}\) | \(\text{100}\) |

The events are whether a bus leaves from Location A or not and whether a bus left late or not.

We test whether the Location A and the left late events are independent.
The total number of buses in the contingency table is \(\text{100}\).
We determine the probabilities of the different events from the values in the table —

- leaving from Location A: \(\frac{40}{100} = \text{0,4}\);
- leaving late: \(\frac{55}{100} = \text{0,55}\);
- leaving from Location A and leaving late: \(\frac{15}{100} = \text{0,15}\).

Since \(\text{0,4} \times \text{0,55} = \text{0,22} \ne \text{0,15}\), the events are dependent.

You are given the following information.

- Events \(A\) and \(B\) are independent.
- \(P(\text{not }A) = \text{0,3}\).
- \(P(B) = \text{0,4}\).

Complete the contingency table below.

| \(A\) | not \(A\) | Totals |

\(B\) | | | |

not \(B\) | | | |

Totals | | | \(\text{50}\) |

From the given table, we see that the total number of outcomes is \(\text{50}\).
Since \(P(\text{not } A) = \text{0,3}\) we have \(n(\text{not } A) = \text{0,3}\times\text{50} = 15\) and \(n(A) = 50 - 15 = 35\).
Since \(P(B) = \text{0,4}\) we have \(n(B) = \text{0,4}\times\text{50} = 20\) and \(n(\text{not } B) = 50 - 20 = 30\).
From this we can partially complete the table:

| \(A\) | not \(A\) | Totals |

\(B\) | | | \(\text{20}\) |

not \(B\) | | | \(\text{30}\) |

Totals | \(\text{35}\) | \(\text{15}\) | \(\text{50}\) |

Next, we use the fact that \(A\) and \(B\) are independent.
From the definition of independence
\[P((\text{not } A) \text{ and } B) = P(\text{not } A) \times P(B) = \text{0,3} \times \text{0,4} = \text{0,12}\]
Therefore \(n((\text{not } A) \text{ and } B) = \text{0,12} \times \text{50} = 6\).
We find the rest of the values in the table by making sure that each row and column sums to its total.

| \(A\) | not \(A\) | Totals |

\(B\) | \(\text{14}\) | \(\text{6}\) | \(\text{20}\) |

not \(B\) | \(\text{21}\) | \(\text{9}\) | \(\text{30}\) |

Totals | \(\text{35}\) | \(\text{15}\) | \(\text{50}\) |