Title: Default for 35.3 Addition and Multiplication Laws of Probability
Course: HELM Workbooks 2023-2024

The addition law

The sample space \(S\) is the set of all possible outcomes of a given experiment. Certain events \(A\) and \(B\) are subsets of \(S\).  Recall what was meant by \(\text{P}(A), \text{P}(B)\) and their complements in the particular case in which the experiment had equally likely outcomes.

Events, like sets, can be combined to produce new events.

- \(A\cup B\) denotes the event that  event \(A\) or event \(B\) (or both) occur when the experiment is performed.
- \(A\cap B\) denotes the event that both \(A\) and \(B\) occur together.

In this quiz we obtain expressions for determining the probabilities of these combined events, which are written \(\text{P}(A\cup B)\) and \(\text{P}(A\cap B)\) respectively.

Types of events

There are two types of events you will need to able to identify and work with: mutually exclusive events and independent events. (We deal with independent events later).

Mutually exclusive events

Mutually exclusive events are events that by definition cannot happen together.  For example, when tossing a coin, the events 'head'  and 'tail'  are mutually exclusive; when testing a switch 'operate' and 'fail' are mutually exclusive; and when testing the tensile strength of a piece of wire, 'hold' and 'snap' are mutually exclusive. In such cases, the probability of both events occurring together must be zero. Hence, using the usual set theory notation for events \(A\) and \(B\), we may write:

\[\qquad \text{P} (A \cap  B)=0,\quad\text{provided that}\ A\ \text{and}\ B\ \text{are mutually exclusive events.}\]

Q: Exercise

Decide which of the following pairs of events (\(A\) and \(B\)) arising from the experiments described are mutually exclusive.

1. Two cards are drawn from a pack:
\(A=\) a red card is drawn.
\(B=\) a picture card is drawn.
\(A\) and \(B\) are [...]
2. The daily traffic accidents in Edinburgh involving pedal cyclists and motorcyclists are counted.
\(A=\) three motorcyclists are injured in collisions with cars.
\(B=\) one pedal cyclist is injured when hit by a bus.
\(A\) and \(B\) are [...]
3. A box contains \(20\) nuts. Some have a metric thread, some have a British Standard Fine (BSF) thread and some have a  British Standard Whitworth (BSW) thread.
\(A=\) first nut picked out of the box is BSF.
\(B=\) second nut picked out of the box is metric.
\(A\) and \(B\) are [...]
A: Not Mutually Exclusive; Mutually Exclusive; Not Mutually Exclusive

Key Point

The Addition Law of Probability - Simple Case

If two events \(A\) and \(B\) are mutually exclusive then
\[\text{P}(A\cup B)=\text{P}(A)+\text{P}(B).\]

Key Point

The Addition Law of Probability - General Case

If two events are \(A\) and \(B\) then

\[\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B).\]

If \(A\cap B= \emptyset\), i.e. \(A\) and \(B\) are mutually exclusive, then \(\text{P}(A\cap B)=\text{P}(\emptyset )=0\), and this general expression reduces to the simpler case.

This rule can be extended to three or more events, for example:

\[\text{P}(A\cup B\cup C)= \text{P}(A)+\text{P}(B)+\text{P}(C) - \text{P}(A\cap B)-\text{P}(A\cap C) -\text{P}(B\cap C)+\text{P}(A\cap B\cap C)\]

Example

Consider a pack of \(52\) playing cards. A card is selected at random.  What is the probability that the card is either a diamond or a ten?

Solution

If \(A\) is the event a diamond is selected and \(B\) is the event a ten is selected then obviously \(\text{P}(A)=\dfrac{13}{52}\) and \(\text{P}(B)=\dfrac{4}{52}\).

The intersection event \(A\cap B\) consists of only one member - the ten of diamonds - which gets counted twice hence \(\text{P}(A\cap B)=\dfrac{1}{52}\).

Therefore \(\text{P}(A\cup B)=\dfrac{13}{52}+\dfrac{4}{52}-\dfrac{1}{52}=\dfrac{16}{52}\).

Exercise

A bag contains \(20\) balls, \([value]\) are coloured red, \([value]\) are coloured green, \([value]\) are coloured blue, \([value]\) are coloured white and \([value]\) are coloured yellow. One ball is selected at random. Find the probabilities of the following events.

1. The ball is either red or green.
2. The ball is not blue.
3. The ball is either red or white or blue.  (Hint: consider the complementary event.)
Feedback: A bag contains \(20\) balls, \([value]\) are coloured red, \([value]\) are coloured green, \([value]\) are coloured blue, \([value]\) are coloured white and \([value]\) are coloured yellow. Note that a ball has only one colour, designated by the letters \(R,G,B,W,Y\). 1. \(\text{P}(R\cup G)=\text{P}(R)+\text{P}(G) = \dfrac{[value]}{20}+\dfrac{[value]}{20} = \displaystyle [value].\) 2. \(\text{P}(B')=1-\text{P}(B)=1-\dfrac{[value]}{20} = \displaystyle [value].\) 3. The complementary event is \(G\cup Y\), \(\text{P}(G\cup Y) = \dfrac{[value]}{20}+\dfrac{[value]}{20}.\) Hence \(\text{P}(R\cup W\cup B)=1-\text{P}(G\cup Y) = \displaystyle [value]\). In part c) we could alternatively have used an obvious extension of the law of addition for mutually exclusive events: \[\text{P}(R\cup W\cup B)=\text{P}(R)+\text{P}(W)+\text{P}(B)=\frac{[value]}{20}+\frac{[value]}{20}+\frac{[value]}{20}=[value].\]

Exercise

The diagram shows a simplified circuit in which two independent components \(a\) and \(b\) are connected in parallel.

[Image]

The circuit functions if either or both of the components are operational. It is known that if \(A\) is the event 'component \(a\) is operating' and \(B\) is the event 'component \(b\) is operating' then \(\text{P}(A)=0.[value]\), \(\text{P}(B)=0.[value]\) and \(\text{P}(A\cap B)=0.[value]\).  Find the probability that the circuit is functioning.
Feedback: The probability that the circuit is functioning is \(\text{P}(A\cup B)\). In words: either \(a\) or \(b\) or both must be functioning if the circuit is to function. Using the keypoint: \[\begin{eqnarray*}\text{P}(A\cup B)&=&\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)\\ &=&0.[value]+0.[value]-0.[value]=0.[value] \end{eqnarray*}\]

Exercise

The following people are in a room: \([value]\) are Spanish aged \(21\) and over, \([value]\) are Spanish under \(21\), \([value]\) are Italian aged \(21\) and over, and \([value]\) are Italian under \(21\).

One person is chosen at random. The following events are defined:

\(A = \{\)the person is aged \(21\) and over\(\};\)
\(B = \{\)the person is under \(21\)\(\}\);
\(C = \{\)the person is Spanish\(\};\)
\(D =\{\) the person is Italian\(\}.\)

Evaluate the following:

1. \(\text{P}(B \cup D)=\)
2. \(\text{P}(A' \cap  C')=\)
Feedback: People in the room: \([value]\) are Spanish aged \(21\) and over, \([value]\) are Spanish under \(21\), \([value]\) are Italian aged \(21\) and over, and \([value]\) are Italian under \(21\). One person is chosen at random. The following events are defined: \(A = \{\)the person is aged \(21\) and over\(\};\) \(B = \{\)the person is under \(21\)\(\}\); \(C = \{\)the person is Spanish\(\};\) \(D =\{\) the person is Italian\(\}.\) 1. We know that \(\text{P}(B\cup D)=\text{P}(B)+\text{P}(D)-\text{P}(B\cap D)\) We calculate: \(\text{P}(B)=\dfrac{[value]+[value]}{18},\ \ \text{P}(D)=\dfrac{[value]+[value]}{18},\ \ \text{P}(B\cap D)=\dfrac[value]{18}\) \[\therefore \quad \text{P}(B\cup D)=\text{P}(B)+\text{P}(D)-\text{P}(B\cap D)=[value].\] 2. \( A'=\{\text{people under}\ 21\}\) \(C'=\{\text{people who are not Spanish (i.e. are Italian)}\}\) \[\therefore\qquad \text{P}(A'\cap C')={[value]\over 18}=[value].\]

Exercise

1. A card is drawn at random from a deck of \(52\) playing cards.  What is the probability that it is an ace or a picture card (i.e.  K, Q, J)?
2. In a single throw of two 6-faced dice, what is the probability that neither a double nor a sum of \(9\) will appear?
Feedback: 1. \[F=\{\text{face card}\}\ \ \ A=\{\text{card is ace}\}\ \ \ \text{P}(F)={12\over 52},\ \text{P}(A)={4\over 52}\] \[\therefore\qquad \text{P}(F\cup A)=\text{P}(F)+\text{P}(A)-\text{P}(F\cap A)={12\over 52}+{4\over 52}-0={16\over 52}.\] 2. \[D=\{\text{double is thrown}\}\qquad N=\{\text{sum is}\ 9\}\] \(\text{P}(D)={6\over 36}\) (\(36\) possible outcomes in an experiment in which all the outcomes are equally probable). \[\text{P}(N)=\text{P}\{(6\cap 3)\cup (5\cap 4)\cup (4\cap 5)\cup (3\cap 6)\}={4\over 36}\] \[\text{P}(D\cup N)=\text{P}(D)+\text{P}(N)-\text{P}(D\cap N)={6\over 36}+{4\over 36}-0={10\over 36}\] \[\text{P}(({D\cup N})')=1-\text{P}(D\cup N)=1-{10\over 36}={26\over 36}.\]

Conditional Probability - Dependent Events

Suppose a bag contains \(6\) balls, \(3\) red and \(3\) white.  Two balls are chosen (without replacement) at random, one after the other. Consider the two events \(R,\ W\):

\(R\) is event "first ball chosen is red''

\(W\) is event "second ball chosen  is white''

We easily find \(\text{P}(R)=\dfrac{3}{6}=\dfrac{1}{2}\).  However, determining the probability of \(W\) is not quite so straightforward.  If the first ball chosen is red then the bag subsequently contains \(2\) red balls and \(3\) white.  In this case \(\text{P}(W)=\dfrac{3}{5}\).  However, if the first ball chosen is white then the bag subsequently contains \(3\) red balls and \(2\) white.  In this case \(\text{P} (W)=\dfrac{2}{5}\).  What this example shows is that the probability that \(W\) occurs is clearly dependent upon whether or not the event \(R\) has occurred.  The probability of \(W\) occurring is conditional on the occurrence or otherwise of \(R\).

The conditional probability of an event \(B\) occurring given that event \(A\) has occurred is written \(\text{P}(B|A)\).  In this particular example

\[\text{P}(W|R)=\dfrac{3}{5}\qquad \text{and}\qquad \text{P}(W|R')=\dfrac{2}{5}.\]

Consider, more generally, the performance of an experiment in which the outcome is a member of an event \(A\). We can therefore say that the event \(A\) has occurred. What is the probability that \(B\) then occurs?  That is what is \(\text{P}(B|A)\)? In a sense we have a new sample space which is the event \(A\).  For \(B\) to occur some of its members must also be members of event \(A\).  So, for example, in an equi-probable space, \(\text{P}(B|A)\) must be the number of outcomes in \(A\cap  B\) divided by the number of outcomes in \(A\).  That is

\[\text{P}(B|A)=\dfrac{\text{number of outcomes in}\ \{A\cap B\}}{\text{number of outcomes in }\ \{A\}}.\]

Now if we divide both the top and bottom of this fraction by the total number of outcomes of the experiment we obtain an expression for the conditional probability of \(B\) occurring given that \(A\) has occurred:

Key Point

Conditional Probability

\[\text{P}(B|A)=\dfrac{\text{P}(A\cap B)}{\text{P}(A)}\qquad \text{or, equivalently}\qquad \text{P}(A\cap B)=\text{P}(B|A)\text{P}(A).\]

To illustrate the use of conditional probability concepts we return to the example of the bag containing \(3\) red and \(3\) white balls in which we consider two events:

- \(R\) is event "first ball is red'';
- \(W\) is event "second ball is white''.

Let the red balls be numbered \(1\) to \(3\) and the white balls \(4\) to \(6\).  If, for example, \((3,5)\) represents the fact that the first ball is \(3\) (red) and the second ball is \(5\) (white) then we see that there are \(6\times 5=30\) possible outcomes to the experiment (no ball can be selected twice).

If the first ball is red then only the fifteen outcomes \((1,x),\ (2,y),\ (3,z)\) are then possible (here \(x\ne 1\), \(y\ne 2\)  and \(z\ne 3\)).  Of these fifteen, the six outcomes \(\{(1,2),\ (1,3),\ (2,1),\ (2,3),\ (3,1),\ (3,2)\}\) will produce the required result, i.e. the event in which both balls chosen are red, giving a probability: \(\text{P}(B|A)=\dfrac{6}{15}=\dfrac{2}{5}\).

Example

A box contains six \(10 \Omega\) resistors and ten \(30 \Omega\) resistors. The resistors are all unmarked and are of the same physical size.

1. One resistor is picked at random from the box; find the probability that:
  1. It is a \(10 \Omega\) resistor.
  2. It is a \(30 \Omega\) resistor.
2. At the start, two resistors are selected from the box. Find the probability that:
  1. Both are \(10 \Omega\) resistors.
  2. The first is a \(10 \Omega\) resistor and the second is a \(30 \Omega\) resistor.
  3. Both are \(30 \Omega\) resistors.

Solution

  1. As there are six \(10 \Omega\) resistors in the box that contains a total of \(6 + 10 = 16\) resistors, and there is an equally likely chance of any resistor being selected, then
\[\text{P}(10\Omega) =\dfrac{6}{16}=\dfrac{3}{8}.\]
  2. As there are ten \(30 \Omega\) resistors in the box that contains a total of \(6 + 10 = 16\) resistors, and there is an equally likely chance of any resistor being selected, then
\[\text{P}(30\Omega) = \dfrac{10}{16}=\dfrac{5}{8}.\]
  1. As there are six \(10 \Omega\) resistors in the box that contains a total of \(6 + 10 = 16\) resistors, and there is an equally likely chance of any resistor being selected, then
\[\text{P(first selected is a }10\Omega \text{ resistor})=\dfrac{6}{16}=\dfrac{3}{8}.\]
If the first resistor selected was a \(10\Omega\) one, then when the second resistor is selected, there are only five \(10\Omega\) resistors left in the box which now contains \(5 + 10 = 15\) resistors. Hence, \[\text{P(second selected is also a } 10 \Omega \text{ resistor}) = \dfrac{5}{15}=\dfrac{1}{3}.\] And, \[\text{P(both are } 10\Omega \text{ resistors})= \dfrac{3}{8}\times\dfrac{1}{3}=\dfrac{1}{8}.\]
  2. As before,
\[\text{P(first selected is a }10\Omega \text{ resistor})=\dfrac{6}{18}=\dfrac{3}{8}.\]
If the first resistor selected was a \(10\Omega\) one, then when the second resistor is selected, there are still ten \(30\Omega\) resistors left in the box which now contains \(5 + 10 = 15\) resistors. Hence,
\[\text{P(second selected is a }30\Omega \text{ resistor})=\dfrac{10}{15}=\dfrac{2}{3}.\] And, \[\text{P(first was a }10\Omega \text{ resistor and second was a }30 \Omega \text{ resistor})=\dfrac{3}{8}\times\dfrac{2}{3}=\dfrac{1}{4}.\]
  3. As there are ten \(30\Omega\) resistors in the box that contains a total of \(6 + 10 = 16\) resistors, and there is an equally likely chance of any resistor being selected, then
\[\text{P(first selected is a }30\Omega \text{ resistor}) = \dfrac{10}{16}\times\dfrac{5}{8}.\]
If the first resistor selected was a \(30\Omega\) one, then when the second resistor is selected, there are only nine \(30\Omega\) resistors left in the box which now contains \(5 +10 =\) 15 resistors.
Hence, \[\text{P(second selected is also a }30\Omega \text{ resistor}) =\dfrac{9}{15}=\dfrac{3}{5}.\]
And,\[\text{P(both are }30\Omega \text{ resistors}) = \dfrac{5}{8}\times\dfrac{3}{5}=\dfrac{3}{8}.\]

Independent Events

If the occurrence of one event \(A\) does not affect, nor is affected by, the occurrence of another event \(B\) then we say that \(A\) and \(B\) are independent events. Clearly, if \(A\) and \(B\) are independent then

\[\text{P}(B|A)=\text{P}(B)\qquad\text{and}\qquad \text{P}(A|B)=\text{P}(A).\]

Then, using the formula \(\text{P}\)\((A\cap B)=\)\(\text{P}\)\((B|A)\text{P}(A)\) we have, for independent events:

Key Point

The Multiplication Law

If \(A\) and \(B\) are independent events then

\[\text{P}(A\cap B)=\text{P}(A)\times \text{P}(B)\]

In words "the probability of independent events \(A\) and \(B\) occurring is the product of the probabilities of the events occurring separately".

In the figure below two components \(a\) and \(b\) are connected in series.

[Image]

Components connected in series

Define two events

- \(A\) is the event 'component \(a\) is operating'.
- \(B\) is the event 'component \(b\) is operating'.

Previous testing has indicated that \(\text{P}(A)=0.99,\) and \(\text{P}(B)=0.98\). The circuit functions only if \(a\) and \(b\) are both operating simultaneously. The components are assumed to be independent.

Then the probability that the circuit is operating is given by

\[\text{P}(A\cap B)=\text{P}(A) \text{P}(B)=0.99\times 0.98=0.9702.\]

Note that this probability is smaller then either \(\text{P}(A)\) or \(\text{P}(B)\).

Q: Exercise

Decide which of the following pairs of events (\(A\) and \(B\)) arising from the experiments described are independent.

1. One card is drawn from each of two packs.
\(A=\) a red card is drawn from pack \(1\).
\(B=\) a picture card is drawn from pack \(2\).
\(A\) and \(B\) are [...]
2. The daily traffic accidents in Hull involving pedal cyclists and motorcyclists are counted.
\(A=\) three motor cyclists are injured in separate collisions with cars.
\(B=\) one pedal cyclist is injured when hit by a bus.
\(A\) and \(B\) are [...]
3. Two boxes contains 20 nuts each, some have a metric thread, some have a British Standard Fine (BSF) threads and some have a British Standard Whitworth (BSW) thread. A nut is picked out of each box.
\(A=\) a nut picked out of the first box is BSF.
\(B=\) a nut picked out of the second box is metric.
\(A\) and \(B\) are [...]
4. A box contains 20 nuts, some have a metric thread, some have a British Standard Fine (BSF) threads and some have British Standard Whitworth (BSW) thread. Two nuts are picked out of the box.
\(A=\) first nut picked out of the box is BSF.
\(B=\) second nut picked out of the box is metric.
\(A\) and \(B\) are [...]
A: Independent Events; Independent Events; Independent Events; Dependent Events

Key Point

Laws of Elementary Probability

Let a sample space \(S\) consist of the \(n\) simple distinct events \(E_1, E_2\dots E_n\) and let \(A\) and \(B\) be events contained in \(S\).

Then:

1. \(0\le \text{P} (A)\le 1\). \(\text{P} (A)=0\) is interpreted as meaning that the event \(A\) cannot occur and \(\text{P} (A)=1\) is interpreted as meaning that the event \(A\) is certain to occur.
2. \(\text{P} (A)+\text{P} (A')=1\) where the event \(A'\) is the complement of the event \(A\).
3. \(\text{P} (E_1)+ \text{P} (E_2)+ \dots + \text{P} (E_n)=1\) where \(E_1, E_2, \dots E_n\) form the sample space.
4. If \(A\) and \(B\) are any two events then \(\text{P} (A\cup B)=\text{P} (A)+\text{P} (B)- \text{P} (A\cap B)\).
5. If \(A\) and \(B\) are two mutually exclusive events then \(\text{P} (A\cup B)=\text{P} (A)+\text{P} (B)\).
6. If \(A\) and \(B\) are two independent events then \(\text{P} (A\cap B)=\text{P} (A)\times \text{P} (B)\).

Example

A circuit has three independent switches \(A\), \(B\) and \(C\) wired in parallel as shown in the figure below.

[Image]

Switches wired in Parallel

Current can only flow through the bank of switches if at least one of them is closed.  The probability that any given switch is closed is \(0.9\).  Calculate the probability that current can flow through the bank of switches.

Solution

Assume that \(A\) is the event {switch \(A\) is closed}. Similarly for switches \(B\) and \(C\). We require \(\text{P} (A\cup B\cup C)\), the probability that at least one switch is closed. Using set theory,

\[\begin{eqnarray*}\text{P} (A\cup B\cup C)&=&\text{P}(A)+\text{P} (B)+\text{P} (C)-[\text{P} (A\cap B)+\text{P} (B\cap C)+\text{P} (C\cap A)]+\text{P} (A\cap B\cap C)\end{eqnarray*}.\] Using the fact that the switches operate independently,

\[\text{P} (A\cup B\cup C)=0.9+0.9+0.9-[0.9\times 0.9 + 0.9\times 0.9+0.9\times0.9]+0.9\times 0.9\times 0.9=0.999.\]

Note that the result implies that the system is more likely to allow current to flow than any single switch in the system. This is why replication is built into systems requiring a high degree of reliability such as aircraft control systems.

Alternative solution:

\[\text{P}(A\cup B\cup C)=1-\text{P}(A'\cap B'\cap C')=1-0.1\times 0.1\times 0.1=0.999.\]

Exercise

A circuit has four independent switches \(A, B, C\) and \(D\) wired in parallel as shown below.

[Image]

Switches in Parallel

Current can only flow through the bank of switches if at least one is closed.  The probabilities that \(A, B, C\) and \(D\)  are closed are \(\displaystyle{[value], [value], [value]}\) and \(\displaystyle{[value]}\) respectively. Calculate the probability that current can flow through the bank of switches.

Hint: First consider the opposite case that the current does not flow.
Feedback: Denoting the switches by \(A, B, C\) and \(D\) we have: \[\begin{eqnarray*}\text{P}(A\cup B\cup C\cup D)&=&\text{P} (A)+\text{P}(B)+\text{P} (C)+\text{P} (D)-\text{P} (A\cap B)- \text{P} (B\cap C)\\&-& \text{P} (C\cap D- \text{P} (D\cap A)- \text{P} (A\cap C)-\text{P} (B\cap D)\\&+&\text{P} (A\cap B\cap C)+ \text{P} (B\cap C\cap D)+\text{P} (C\cap D\cap A)+\text{P} (D\cap A\cap B)\\&-& \text{P} (A\cap B\cap C\cap D)\end{eqnarray*}\] Using the fact that the switches operate independently and substituting gives: \[ \text{P} (A\cup B\cup C\cup D)=3-3.35+1.65-0.3024=0.9976\] Hence, the probability that current can flow through the bank of switches is \(\displaystyle [value]\). Alternative solution: Consider the opposite case where the current does not flow through, then all of the switches need to be open. If \(A\), \(B\), \(C\) and \(D\) represent the event that the respective switch is closed, then the event \(A’ \cap B’ \cap C’ \cap D’\) represents all four switches being closed. As each switch is independent, we have: \[ P(A’\cap B’\cap C’\cap D’) = P(A’)P(B’)P(C’)P(D’) = [value]\times[value]\times[value]\times[value]= \frac{3}{1250}. \] Therefore the probability that the current flows is \begin{eqnarray*}\text{P}(A\cup B\cup C\cup D)&=&1-\text{P}(A'\cap B'\cap C'\cap D')\\&=&1- \frac{3}{1250}=0.9976.\end{eqnarray*}

Exercise

1. A box contains \([value]\) bad tube(s) and \([value]\) good tubes.  Two are drawn out together.  One of them is tested and found to be good.  What is the probability that the other one is also good?
2. A man owns a house in town and a cottage in the country.  In any one year the probability of the town house being burgled is \(\displaystyle{[value]}\) and the probability of the country cottage being burgled is\(\displaystyle{[value]}\).  In any one year what is the probability that:
(1) both will be burgled?
(2) one or the other (but not both) will be burgled?
3. In a Baseball Series, teams \(A\) and \(B\) play until one team has won \([value]\) games.  If team \(A\) has probability \(\dfrac{2}{3}\) of winning against \(B\) in a single game, what is the probability that the Series will end only after \([value]\) games are played?
Feedback: 1. Let \(G_i=\{i^{th} \text{ tube is good}\}\quad B_i=\{i^{th} \text{ tube is bad}\}\) \(\text{P}(G_2|G_1)=[value]\) (only \(([value]-1)\) good tubes left out of \(9\)). 2. (1) \(H=\{\)house is burgled\(\}\ \ C=\{\)cottage is burgled\(\}\) \[\text{P}(H\cap C)=\text{P}(H)\text{P}(C)=\left([value]\right)\left([value]\right)=[value]\] since events independent. (2) \[\text{P(one or the other (but not both))}=\text{P}((H\cap C')\cup (H' \cap C))=\text{P}(H\cap C')+\text{P}(H'\cap C)\]\[=\text{P}(H)\text{P}(C')+\text{P}(H')\text{P}(C)=\left([value]\right)\left([value]\right)+\left([value]\right)\left([value]\right)=[value].\] 3. Let \(A_i\) be event \(\{A\) wins the \(i^{th}\) game\(\}\). The required event is \(A\) winning \([value]-1\) games and losing \([value]\) games in any order. \[\text{P(required event)}=\binom{[value]-1}{[value]-1}[\text{P}(A_1)]^{[value]-1}[\text{P}(A'_1)]^{[value]}=[value].\]

Exercise

The probability that a single aircraft engine will fail during flight is \(q\). A multi-engine plane makes a successful flight if at least half its engines run. Assuming that the engines operate independently, find the values of \(q\) for which a two-engine plane is to be preferred to a four-engine plane.

\(<q<\)

Hint: Consider how many different ways could you have two/three/four engines still working on a four-engine plane.
Feedback: Let \(E_i\) be event \(\{i^{th}\) engine success\(\}\) Two-engine plane: flight success if \(\{(E_1\cap E_2)\cup ({E'}_1\cap E_2)\cup(E_1\cap {E'}_2)\}\) occurs. \[\begin{eqnarray*}\text{P}\text{(required event)}\ &=&\text{P}(E_1)\text{P}(E_2)+\text{P}(E'_1)\text{P}(E_2)+\text{P}(E_1)\text{P}(E'_2)\\ &=&(1-q)^2+2q(1-q)=1-q^2\end{eqnarray*}\] Four-engine plane: success if following event occurs \[\underbrace{\{E_1\cap E_2\cap E'_3\cap E'_4\}}_{^4C_2\ \text{ways}} \cup \underbrace{\{E_1\cap E_2\cap E_3\cap E'_4\}}_{^4C_1\ \text{ways}}\ \cup \underbrace{\{E_1\cap E_2\cap E_3\cap E_4\}}_{^4C_0\ \text{ways}}\] required probability\(= \ 6(1-q)^2q^2+4(1-q)^3q+(1-q)^4=3q^4-4q^3+1\) Two-engine plane is preferred if \[1-q^2>3q^4-4q^3+1\qquad \text{i.e. if}\quad 0>q^2(3q-1)(q-1).\] Let \(y=(3q-1)(q-1)\). By drawing a graph of this quadratic you will quickly see that a two-engine plane is preferred if \({1\over 3}<q<1\).

Exercise

A box contains \([value]\) bad tube(s) and \([value]\) good tubes. Two are drawn out together. One of them is tested and found to be good. What is the probability that the other one is also good?
Feedback: A box contains \([value]\) bad tube(s) and \([value]\) good tubes, so, in total [value] tubes. Let \(G_i=\{i^{th} \text{ tube is good}\}.\) The probability to draw one good tube out of the box is \( G_1 = \displaystyle [value]. \) If the first tube was good, then when the second tube is selected there are only [value] good tubes left in the box which now contains [value] tubes. Hence, \(\text{P}(G_2|G_1)= \displaystyle [value].\)

Q: Exercise

Decide which of the following pairs of events (\(A\) and \(B\)) arising from the experiments described are mutually exclusive.

1. Two cards are drawn from a pack:
\(A=\) a red card is drawn.
\(B=\) a picture card is drawn.
\(A\) and \(B\) are [...]
2. A participant is recruited to a clinical trial who is randomly assigned to receive one of two treatments:
\(A=\) the participant receives a placebo.
\(B=\) the participant receives the experimental treatment.
\(A\) and \(B\) are [...]
3. A box contains \(20\) nuts. Some have a metric thread, some have a British Standard Fine (BSF) thread and some have a  British Standard Whitworth (BSW) thread.
\(A=\) first nut picked out of the box is BSF.
\(B=\) second nut picked out of the box is metric.
\(A\) and \(B\) are [...]
A: Not Mutually Exclusive; Mutually Exclusive; Not Mutually Exclusive
Feedback: 1. It is possible to draw a red picture card (e.g. Jack of hearts) meaning that events \(A\) and \(B\) are not mutually exclusive. 2. A participant is randomly assigned to either the placebo or the experimental treatment, never both. This means that events \(A\) and \(B\) are mutually exclusive. 3. As the box contains at least one of the three types if nuts, there is a non-zero probability of selecting one that is BSF and another that is metric. Therefore events \(A\) and \(B\) are not mutually exclusive.

Q: Exercise

Decide which of the following pairs of events (\(A\) and \(B\)) arising from the experiments described are independent.

1. One card is drawn from each of two packs.
\(A=\) a red card is drawn from pack \(1\).
\(B=\) a picture card is drawn from pack \(2\).
\(A\) and \(B\) are [...]
2. A manufacture of cutting tools sources synthetic diamonds from two laboratories:
\(A=\) the quality of the diamonds sourced from lab A meet the required specifications.
\(B=\) the quality of the diamonds sourced from lab B meet the required specifications.
\(A\) and \(B\) are [...]
3. Two boxes contains 20 nuts each, some have a metric thread, some have a British Standard Fine (BSF) threads and some have a British Standard Whitworth (BSW) thread. A nut is picked out of each box.
\(A=\) a nut picked out of the first box is BSF.
\(B=\) a nut picked out of the second box is metric.
\(A\) and \(B\) are [...]
4. A box contains 20 nuts, some have a metric thread, some have a British Standard Fine (BSF) threads and some have British Standard Whitworth (BSW) thread. Two nuts are picked out of the box.
\(A=\) first nut picked out of the box is BSF.
\(B=\) second nut picked out of the box is metric.
\(A\) and \(B\) are [...]
A: Independent Events; Independent Events; Independent Events; Dependent Events
Feedback: 1. \(P(A) = \frac{1}{4}\), \(P(B) = \frac{3}{13}\) and \(P(A \cap B) = \frac{3}{52}\). \(P(A\cap B) = P(A)P(B)\) so \(A\) and \(B\) are independent events. 2. The process of growing diamonds by lab A does not influence how lab B grows diamonds, and visa versa. So events \(A\) and \(B\) are independent. 3. The nuts are sampled from different boxes, so \(A\) and \(B\) are independent events. 4. The nuts are sampled from the same box without replacement. The means that the probability that the second nut is metric will differ if the first sampled nut was or was not a metric nut. Therefore, events \(A\) and \(B\) are dependent.

Exercise

A man owns a house in town and a cottage in the country. In any one year the probability of the town house being burgled is \(\displaystyle{[value]}\) and the probability of the country cottage being burgled is\(\displaystyle{[value]}\).

In any one year what is the probability that:

1. Both will be burgled?
2. One or the other (but not both) will be burgled?
Feedback: A man owns a house in town and a cottage in the country. In any one year the probability of the town house being burgled is \(\displaystyle{[value]}\) and the probability of the country cottage being burgled is\(\displaystyle{[value]}\). 1. \(H=\{\)house is burgled\(\}\ \ C=\{\)cottage is burgled\(\}\) \[\text{P}(H\cap C)=\text{P}(H)\text{P}(C)=\left([value]\right)\left([value]\right)=[value]\] since events independent. 2. \[\begin{align} \text{P(one or the other (but not both))} & = \text{P}((H \cap C')\cup (H' \cap C)) \\ & =\text{P}(H\cap C')+\text{P}(H'\cap C) \\ &= \text{P}(H)\text{P}(C')+\text{P}(H')\text{P}(C) \\ & = \left([value]\right)\left([value]\right)+\left([value]\right)\left([value]\right) \\ &=[value]. \end{align}\]

Exercise

Current flows through a relay only if it is closed. The probability of any relay being closed is \(\displaystyle{[value]}\). Assuming that all the relays operate independently, calculate the probability that a current will flow through a circuit composed of 3 relays in parallel.

\( P = \)
Feedback: 1. Let \(A\) be event \(\{\)relay \(A\) is closed\(\}\): Similarly for \(B,\ C\). The required event is \[\{A\cap B\cap C\}\cup\underbrace{\{A'\cap B\cap C\}}_{ ^3C_1}\cap \underbrace{\{A'\cap B'\cap C\}}_{ ^3C_2}\] \[\text{P(required event)} = \left([value]\right)^3+3\left([value]\right)^2\left([value]\right)+3\left([value]\right)\left([value]\right)^2=[value]\] It is often easier to think of this as \(1-\text{P(all relays open)} =1-\left([value]\right)^3=[value]\).

Exercise

A central heating installation and maintenance engineer keeps a record of the causes of failure of systems he is called out to repair. The causes of failure are classified as 'electrical', 'gas' or in some cases 'other' . A summary of the records kept of failures involving either gas or electrical faults is as follows:

[Image: 35.3.3 exercises figure]

1. Find the probability that failure involves gas given that it involves electricity.
2. Find the probability that failure involves electricity given that it involves gas.
Feedback: 1. A total of \(76\) failures involved electrical faults. Of the \(76\) some \(53\) involved gas. Hence \[\text{P(Gas Failure | Electrical Failure)}=[value].\] 2. A total of \(64\) failures involved electrical faults. Of the \(64\) some \(53\) involved gas. Hence \[\text{P(Electrical Failure | Gas Failure)}=[value].\]