End of chapter exercises

The experiment is the outcome of selecting a particular model and colour car from the sample space of available cars.

The sample space is \(\{\)pink model \(A\); lime-green model \(A\); purple model \(A\); purple model \(B\); orange model \(B\); multicolour model \(B\)\(\}\).

Since there are only two models, namely \(A\) and \(B\), the probability of stealing one of the two models is \(\text{1}\).

Since there is no overlap between the two models and since the thief steals only one car, the probability of stealing both of the models is \(\text{0}\).

From the addition rule: \begin{align*} P(H\text{ or }J) &= P(H) + P(J) - P(H\text{ and }J) \\ &= \text{0,62} + \text{0,39} - \text{0,31} \\ &= \text{0,7} \end{align*}

We draw a Venn diagram to get an idea of what the event \((H' \text{ or } J')\) looks like.

\(H'\)	\(J'\)	\(H' \text{ or } J'\)

From the third diagram above we can see that \begin{align*} P(H' \text{ or } J') &= 1 - P(H\text{ and }J) \\ &= \text{0,69} \end{align*}

We draw a Venn diagram to get an idea of what the event \((H' \text{ or } J)\) looks like.

\(H'\)	\(J\)	\(H' \text{ or } J\)

From the third diagram above we can see that \begin{align*} P(H' \text{ or } J) &= P(H') + P(H\text{ and }J) \\ &= \text{0,38} + \text{0,31} \\ &= \text{0,69} \end{align*}

We draw a Venn diagram to get an idea of what the event \((H' \text{ and } J')\) looks like.

\(H'\)	\(J'\)	\(H' \text{ and } J'\)

From the third diagram above we can see that \begin{align*} P(H' \text{ and } J') &= 1 - P(H\text{ or }J) \\ &= \text{0,3} \end{align*}

The Venn diagram is shown below. (Note that Y is not a vowel.)

We can then read off all the required probabilities.

\(P(\text{not } F) = 1 - \frac{4}{10} = \frac{3}{5}\)

This leaves only the outcomes in \(D\), so \(P(\text{neither } E \text{ nor } F) = \frac{1}{10}\).

Since \(D\) and \(E\) are complementary (a letter is either a vowel or a consonant), \(D' = E\), so \(P(E \text{ and } D') = P(E \text{ and } E) = P(E) = \frac{9}{10}\).

We first draw the outline of the Venn diagram.

Next, we determine the counts of the different regions in the diagram, using the information provided.

• There are \(\text{7}\) people who like all three neighbourhoods, so \(n(A\text{ and }B\text{ and }C) = 7\).
• There are \(\text{21}\) people who like \(A\) and \(C\), so \(n(A\text{ and }C) = 21\).
• There are \(\text{18}\) people who like \(B\) and \(C\), so \(n(B\text{ and }C) = 18\).
• There are \(\text{40}\) people who like \(C\), so \(n(C) = 40\).
• Since \(\text{68}\) people like at least one neighbourhood and since there are \(\text{80}\) people in total, \(\text{12}\) people like none of the neighbourhoods.

Using the above information we can complete part of the Venn diagram.

Since \(n(A\text{ or }B\text{ or }C) = 68\), we can see from the diagram above that \(n(A\text{ or }B) = 60\). From the sum rule \[n(A\text{ and }B) = n(A) + n(B) - n(A\text{ or }B) = 40 + 35 - 60 = 15\] This allows us to complete the diagram:

\(G\) and \(H\) are mutually exclusive when \(P(G \text{ and } H) = 0\).

From the sum rule we have \[P(G \text{ or } H) = P(G) + P(H) - P(G \text{ and } H)\] By substituting in the given values and \(P(G \text{ and } H) = 0\), we get \[\text{0,7} = \text{0,4} + h - 0\] Therefore \(h = \text{0,3}\).

\(G\) and \(H\) are independent when \(P(G \text{ and } H) = P(G) \times P(H)\).

From the sum rule we have \[P(G \text{ or } H) = P(G) + P(H) - P(G \text{ and } H)\] By substituting in the given values and \(P(G \text{ and } H) = P(G) \times P(H)\), we get \begin{align*} \text{0,7} &= \text{0,4} + h - \text{0,4} \times h \\ \text{0,3} &= \text{0,6} \times h \end{align*} Therefore \(h = \text{0,5}\).

The outcomes where Team 1 wins (when the first score is greater than the second score) are underlined in the tree diagram below.

We compute the probability of Team 1 winning by multiplying the probabilities along each path and adding them up. \begin{align*} \text{0,52} \times \text{0,65} \times \text{0,75} &= \text{0,2535} \\ \text{0,52} \times \text{0,65} \times \text{0,25} &= \text{0,0845} \\ \text{0,48} \times \text{0,4} \times \text{0,5} &= \text{0,096} \\ \\ \text{0,2535} + \text{0,0845} + \text{0,096} &= \underline{\text{0,434}} \end{align*}

The outcomes where the game is a draw (when the first score equals the second score) are underlined in the tree diagram below.

There is only one such outcome and so the probability of a draw is the product of the probabilities along the path. \[\text{0,52} \times \text{0,35} = \text{0,182}\]

The outcomes where the sum of the points is even are underlined in the tree diagram below.

\begin{align*} \text{0,52} \times \text{0,35} &= \text{0,182} \\ \text{0,48} \times \text{0,6} &= \text{0,288} \\ \\ \text{0,182} + \text{0,288} &= \underline{\text{0,47}} \end{align*}