Independent events

3 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

$P (B | A) = P (B) and P (A | B) = P (A)$

Then, using the Key Point 7 formula $P (A \cup B) = P (B | A) P (A)$ we have, for independent events:

Key Point 8

The Multiplication Law

If $A$ and $B$ are independent events then

P (A \cap B) = P (A) \times 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 Figure 8 two components $a$ and $b$ are connected in series.

Figure 8

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Define two events

$A$ is the event ‘component $a$ is operating’
$B$ is the event ‘component $b$ is operating’

Previous testing has indicated that $P (A) = 0.99,$ and $P (B) = 0.98$ . 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

$P (A \cap B) = P (A) P (B) = 0.99 \times 0.98 = 0.9702$

Note that this probability is smaller then either $P (A)$ or $P (B)$ .

Task!

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

One card is drawn from each of two packs
$A = {a red card is drawn from pack 1}$

$B = {a picture card is drawn from pack 2}$
The daily traffic accidents in Hull involving pedal cyclists and motor cyclists are counted
$A = {three motor cyclists are injured in separate collisions with cars}$

$B = {one pedal cyclist is injured when hit by a bus}$
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.
$A = {nut picked out of the first box is BSF}$

$B = {nut picked out of the second box is metric}$
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.
$A = {first nut picked out of the box is BSF}$

$B = {second nut picked out of the box is metric}$

(1), (2), (3): $A$ and $B$ are independent. (4) $A$ and $B$ are not independent.

Key Point 9

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:

$0 \leq P (A) \leq 1$ . $P (A) = 0$ is interpreted as meaning that the event $A$ cannot occur and $P (A) = 1$ is interpreted as meaning that the event $A$ is certain to occur.
$P (A) + P (A^{'}) = 1$ where the event $A^{'}$ is the complement of the event $A$
$P (E_{1}) + P (E_{2}) + \dots + P (E_{n}) = 1$ where $E_{1}, E_{2}, \dots E_{n}$ form the sample space
If $A$ and $B$ are any two events then $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
If $A$ and $B$ are two mutually exclusive events then $P (A \cup B) = P (A) + P (B)$
If $A$ and $B$ are two independent events then $P (A \cap B) = P (A) \times P (B)$ .

Example 12

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

Figure 9

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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 $P (A \cup B \cup C)$ , the probability that at least one switch is closed. Using set theory,

\begin{array}{rcl} P (A \cup B \cup C) & = & P (A) + P (B) + P (C) - [P (A \cap B) + P (B \cap C) + P (C \cap A)] \\ + P (A \cap B \cap C) \end{array}

Using the fact that the switches operate independently,

\begin{array}{rcl} P (A \cup B \cup C) & = & 0.9 + 0.9 + 0.9 - [0.9 \times 0.9 + 0.9 \times 0.9 + 0.9 \times 0.9] \\ + 0.9 \times 0.9 \times 0.9 \\ = & 2.7 - 2.43 + 0.729 = 0.999 \end{array}

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.

Task!

A circuit has four independent switches $A, B, C$ and $D$ wired in parallel as shown in the diagram below.

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Current can only flow through the bank of switches if at least one of them is closed. The probabilities that switches $A, B, C$ and $D$ are closed are 0.9, 0.8, 0.7 and 0.6 respectively. Calculate the probability that current can flow through the bank of switches.

Denoting the switches by $A, B, C$ and $D$ we have:

$P (A \cup B \cup C \cup D)$

$= P (A) + P (B) + P (C) + P (D)$

$- P (A \cap B) - P (B \cap C) - P (C \cap D) - P (D \cap A) - P (A \cap C) - P (B \cap D)$

$+ P (A \cap B \cap C) + P (B \cap C \cap D) + P (C \cap D \cap A) + P (D \cap A \cap B)$

$- P (A \cap B \cap C \cap D)$

Using the fact that the switches operate independently and substituting gives:

$P (A \cup B \cup C \cup D) = 3 - 3.35 + 1.65 - 0.3024 = 0.9976$

Hence, the probability that current can flow through the bank of switches is 0.9976.

Exercises

A box contains 4 bad tubes and 6 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?
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 0.01 and the probability of the country cottage being burgled is 0.05. 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 ?
In a Baseball Series, teams $A$ and $B$ play until one team has won 4 games. If team $A$ has probability 2/3 of winning against $B$ in a single game, what is the probability that the Series will end only after 7 games are played?
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.
Current flows through a relay only if it is closed. The probability of any relay being closed is 0.95. Calculate the probability that a current will flow through a circuit composed of 3 relays in parallel. What assumption must be made?
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:
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.

Let $G_{i} = {i^{t h} tube is good} B_{i} = {i^{t h} tube is bad}$
$P (G_{2} | G_{1}) = \frac{5}{9}$ (only 5 good tubes left out of 9).
1. $H = {house is burgled}$ $C = {cottage is burgled}$
2. $P (H \cap C) = P (H) P (C) = (0.01) (0.05) = 0.0005$ since events independent
  $P$ (one or the other (but not both)) $= P ((H \cap C^{'}) \cup (H^{'} \cap C)) = P (H \cap C^{'}) + P (H^{'} \cap C)$
  
  $= P (H) P (C^{'}) + P (H^{'}) P (C)$
  
  $= (0.01) (0.95) + (0.99) (0.05) = 0.059 .$
Let $A_{i}$ be event ${A wins the i^{t h} game}$
required event is $\underset{no. of ways of arranging 3 in 6 i.e._{}^{6} C_{3}}{\underset{⏟}{{A_{1} \cap A_{2} \cap A_{3} \cap A_{4}^{'} \cap A_{5}^{'} \cap A_{6}^{'}) \cap (\dots)}}$

$P$ (required event) $=_{}^{6} C_{3} P (A_{1} \cap A_{2} \cap A_{3} \cap A_{4}^{'} \cap A_{5}^{'} \cap A_{6}^{'}) =_{}^{6} C_{3} {[P (A_{1})]}^{3} P {[A_{1}^{'}]}^{3} = \frac{160}{729}$
Let $E_{i}$ be event ${i^{t h} engine success}$
Two-engine plane: flight success if ${(E_{1} \cap E_{2}) \cup ({E^{'}}_{1} \cap E_{2}) \cup (E_{1} \cap {E^{'}}_{2})}$ occurs
$\begin{array}{rcl} P (required event) & = & P (E_{1}) P (E_{2}) + P (E_{1}^{'}) P (E_{2}) + P (E_{1}) P (E_{2}^{'}) \\ = & {(1 - q)}^{2} + 2 q (1 - q) = 1 - q^{2} \end{array}$
Four-engine plane: success if following event occurs

$\underset{^{4} C_{2} ways}{\underset{⏟}{{E_{1} \cap E_{2} \cap E_{3}^{'} \cap E_{4}^{'}}}} \cap \underset{^{4} C_{1} ways}{\underset{⏟}{{E_{1} \cap E_{2} \cap E_{3} \cap E_{4}^{'}}}} \cap \underset{^{4} C_{0} ways}{\underset{⏟}{{E_{1} \cap E_{2} \cap E_{3} \cap E_{4}}}}$

required probability $= 6 {(1 - q)}^{2} q^{2} + 4 {(1 - q)}^{3} q + {(1 - q)}^{4} = 3 q^{4} - 4 q^{3} + 1$

Two-engine plane is preferred if

$1 - q^{2} > 3 q^{4} - 4 q^{3} + 1 i.e. if 0 > q^{2} (3 q - 1) (q - 1)$

Let $y = (3 q - 1) (q - 1)$ . By drawing a graph of this quadratic you will quickly see that a two-engine plane is preferred if $\frac{1}{3} < q < 1$ .
Let $A$ be event ${relay A is closed}$ : Similarly for $B, C$
required event is ${A \cap B \cap C} \cup \underset{^{3} C_{1}}{\underset{⏟}{{A^{'} \cap B \cap C}}} \cap \underset{^{3} C_{2}}{\underset{⏟}{{A^{'} \cap B^{'} \cap C}}}$

$P$ (required event) $= {(0.95)}^{3} + 3 {(0.95)}^{2} (0.05) + 3 (0.95) {(0.05)}^{2} = 0.999875$

( or $1 - P$ (all relays open) $= 1 - {(0.05)}^{3} = 0.999875$ .)

The assumption is that relays operate independently.
1. A total of 76 failures involved electrical faults. Of the 76 some 53 involved gas. Hence
  $P {Gas Failure | Electrical Failure} = \frac{53}{76} = 0.697$
2. A total of 64 failures involved electrical faults. Of the 64 some 53 involved gas. Hence
  $P {Electrical Failure | Gas Failure} = \frac{53}{64} = 0.828$

3 Independent events

Key Point 8

Task!

Answer

Key Point 9

Example 12

Solution

Task!

Answer

Exercises

Answer