Probability: part 1  (Page 6/7)

Probability in everyday life

Probability is connected with uncertainty. In any statistical experiment, the outcomes that occur may be known, but exactly whichone might not be known. Mathematically, probability theory formulates incomplete knowledge related to the likelihood of an occurrence. For example, ameteorologist might say there is a 60% chance that it will rain tomorrow. This means that in 6 of every 10 times when the world is in the current state, itwill rain tomorrow.

Another way of referring to probabilities is odds. The odds of an event is defined as the ratio of the probability that the eventoccurs to the probability that it does not occur. For example, the odds of a coin landing on a given side are $\frac{0.5}{0.5}=1$ , usually written "1 to 1" or "1:1". This means that on average, the coin will land on that side as manytimes as it will land on the other side.

The simplest example: equally likely outcomes

We say two outcomes are equally likely if they have an equal chance of happening. For example when a fair coin is tossed, each outcome in the samplespace $S=\{heads,tails\}$ is equally likely to occur.

Probability is a function of events (since it is not possible to have a single event with two different probabilities occurring), so we usually denote the probability $P$ of some event $E$ occurring by $P\left(E\right)$ . When all the outcomes are equally likely (in any activity), it is fairly straightforward to count the probability of a certain event occuring. In this case,

$P\left(E\right)=n\left(E\right)/n\left(S\right)$

For example, when you throw a fair die the sample space is $S=\{1;2;3;4;5;6\}$ so the total number of possible outcomes $n\left(S\right)=6$ .

Event 1: Get a 4

The only possible outcome is a $4$ , i.e $E=\left\{4\right\}$ . So $n\left(E\right)=1$ .

Probability of getting a 4: $P(k=4)=n\left(E\right)/n\left(S\right)=1/6$ .

Event 2: Get a number greater than 3

Favourable outcomes: $E=\{4;5;6\}$

Number of favourable outcomes: $n\left(E\right)=3$ .

Probability of getting a number greater than 3: $P(k>3)=n\left(E\right)/n\left(S\right)=3/6=1/2$ .