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