Probability

Hong Kong

Stage 1 - Stage 3

Lesson

In many situations, we calculate how likely it is that an event will occur by counting all the possible outcomes and also the number of outcomes that would be considered a success for that event.

We then make a fraction which we call the probability of the event: $\frac{\text{successful outcomes}}{\text{possible outcomes}}$successful outcomespossible outcomes.

In many other situations, it is not possible to estimate probabilities by counting (or measuring) anything. We instead rely on past experience or historical data concerning the outcomes of similar experiments or processes. We assume that if event $A$`A` occurred in $75%$75% of similar situations in the past, the probability of event $A$`A` occurring this time is $75%$75%.

Sometimes, the event that we are interested in is really a combination of two other events. The two events that make up the compound event are connected in ordinary language by the words OR or AND.

For example, a certain sporting event will be CANCELLED if THE CREEK IS FLOODED * or *if IT IS RAINING HEAVILY on the day. This means that if either one of the events connected by the word

In other situations, we might say event $C$`C` occurs whenever both event $A$`A` occurs AND event $B$`B` occurs. For example, you will receive an AWARD if you EXCEED THE TARGET * and *if your TEAM DOES NOT FAIL.

The mathematical issue to be resolved in dealing with compound events is the question of how the *probabilities *of events joined by the words AND or OR are to be combined to give the probability of the compound event.

The problem can be illustrated with diagrams like the following, called Venn diagrams. The ideas behind the diagrams are taken from the theory of sets.

The rectangular box contains all the outcomes that can occur in a process or experiment. The shapes inside the box contain the outcomes that would constitute the occurrence of events $A$`A` and $B$`B`. Notice that some outcomes belong both to events $A$`A` and $B$`B`. This possibility is represented by the overlap of the shapes, which is called the *intersection *of events $A$`A` and $B$`B`, and is notated $A\cap B.$`A`∩`B`.

If we wish to combine the probabilities associated with events $A$`A` and $B$`B` when either $A$`A` or $B$`B` occurs, the natural thing to do would be to just add the probabilities. We would do this if we were thinking of a compound event that occurs when either event $A$`A` or event $B$`B` occurs, or both. In the language of set theory, this combination of events is called the union of $A$`A` and $B$`B`. It is notated $A\cup B$`A`∪`B`.

However, the diagram suggests that if we add the probabilities associated with the two events, then some of the probability will be included twice because of the overlapping portion. We need to compensate for this by making sure that the probability associated with the intersection is only included once.

In words, we would say: The probability of the occurrence of $A$`A` OR $B$`B` is equal to the probability of $A$`A` occurring plus the probability of $B$`B` occurring minus the probability of both occurring. A mathematical shorthand for this is:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$`P`(`A`∪`B`)=`P`(`A`)+`P`(`B`)−`P`(`A`∩`B`)

It can happen that there is no outcome that would lead to the claim that events $A$`A` and $B$`B` had both occurred. In this case, there would be no overlap of events in the Venn diagram. In such a situation, we say events $A$`A` and $B$`B` are *disjoint *events.

In the case of disjoint events $A$`A` and $B$`B`, the addition rule simplifies to

$P(A\cup B)=P(A)+P(B)$`P`(`A`∪`B`)=`P`(`A`)+`P`(`B`)

The overlap of events $A$`A` and $B$`B` in the Venn diagram above, which we called the intersection of $A$`A` and $B$`B`, is the set of outcomes that would enable us to say that both $A$`A` AND $B$`B` had occurred.

To see how the separate probabilities of events $A$`A` and $B$`B` can be combined to give the probability of the intersection event $A\cap B$`A`∩`B`, we need to introduce the concept of *independence *of events.

In everyday language, we mean by independence that the chance of one event occurring is not affected by the occurrence of the other.

The equivalent mathematical way of expressing this idea is to say events $A$`A` and $B$`B` are independent if the probability of event $A$`A` in the whole outcome space is the same as the probability of event $A$`A` occurring given that event $B$`B` occurs. In the Venn diagram, for independence, we require that the proportion of the area representing event $A$`A` in the rectangle representing the whole outcome space is the same as the proportion of the area $A\cap B$`A`∩`B` in the area representing event $B$`B`. This may be easier to see when expressed in mathematical notation:

If events $A$`A` and $B$`B` are independent, $\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{1}$`P`(`A`∩`B`)`P`(`B`)=`P`(`A`)1.

If this condition is fulfilled, we can multiply the separate probabilities to obtain the probability of the intersection. That is, if the occurrences of events $A$`A` and $B$`B` do not influence one another, we can say the probability of $A$`A` AND $B$`B` both occurring is given by

$P(A\cap B)=P(A).P(B)$`P`(`A`∩`B`)=`P`(`A`).`P`(`B`)

The numbers $1$1 to $19$19 are to be drawn randomly from a hat. How many odd numbers are there? How many prime numbers are there? How many are both odd and prime? How many are both even and prime? How many are both even and odd?

There are $10$10 odd numbers: $1,3,5,7,9,11,13,15,17,19$1,3,5,7,9,11,13,15,17,19.

There are $8$8 prime numbers: $2,3,5,7,11,13,17,19$2,3,5,7,11,13,17,19.

Seven numbers are both odd and prime.

One number is even and prime.

No numbers are both odd and even. The odd numbers and the even numbers are disjoint sets.

If the experiment of drawing a number from a hat containing the numbers 1 to 5 is done twice in succession with the number drawn replaced after each trial, list all the possible outcomes as ordered pairs of numbers. In how many outcomes of this experiment are either the first number or the second number or both odd? In how many outcomes are both numbers odd? If the first number is even, in how many outcomes is the second number odd?

We could list the outcomes in a table, as follows:

$1,1$1,1 | $1,2$1,2 | $1,3$1,3 | $1,4$1,4 | $1,5$1,5 |

$2,1$2,1 | $2,2$2,2 | $2,3$2,3 | $2,4$2,4 | $2,5$2,5 |

$3,1$3,1 | $3,2$3,2 | $3,3$3,3 | $3,4$3,4 | $3,5$3,5 |

$4,1$4,1 | $4,2$4,2 | $4,3$4,3 | $4,4$4,4 | $4,5$4,5 |

$5,1$5,1 | $5,2$5,2 | $5,3$5,3 | $5,4$5,4 | $5,5$5,5 |

There are $21$21 cells that contain at least one odd number.

There are $9$9 cells that contain two odd numbers.

There are $6$6 cells in which the first number is even and the second odd.

If we define event $A$`A` to be {*An odd number occurs on the first draw*}, and event $B$`B` to be {*An odd number occurs on the second draw*} we would say that $A\cup B$`A`∪`B` occurs $21$21 times and $A\cap B$`A`∩`B` occurs $9$9 times.