## Basic Probability, part 1

### August 22, 2009

In this part we are going to discuss classical probability and it’s applications.

What does probability mean?

Consider an example. Let’s toss a die. There are only 6 possible outcomes for us: ultimately a die shows upper side with only one number and there are 6 numbers. All these possible outcomes are called sample space. So our sample space is $S=\{1,2,3,4,5,6\}.$

Now let us consider some events: the die shows 1, the die shows 2, the die shows 1 or 2, the dies shows an even number (2 or 4 or 6), and so on. Choose the event – the die shows 2.

The formula for determining probability for an event E is:

$P(E)= \frac{number~ of~ outcomes~ in ~ the ~event}{number~ of~ outcomes~ in ~ the ~ sample~ space}.$

The number of outcomes in the event “the dies shows 2” is 1, because it is only one case when we can get number 2. The number of outcomes in the sample space is the number of all possible cases and, as mentioned upper, is 6.
So, we can write our first probability: the probability of the event that the die shows 2 is $P(E)= \frac{1}{6}.$

The event “the die shows an odd number” has 3 outcomes: 1 or 3 or 5. Then probability for this event will be $P(1 ~or ~3 or~ 5)= \frac{3}{6}=\frac{1}{2}.$

Now let us consider a coin. The sample space is $S=\{Heads, Tails \}.$

The probability that tossed coin shows H(Heads) is $\frac{1}{2}.$ The same is for T(Tails).

And there is more complicated case: let us toss two coins instantly. We have 4 outcomes: HH, HT, TH, TT.
What is the probability of the event that we get H on the first coin and H on the second, or in other words P(HH).

It is equal $\frac{1}{4}$, since we have 1 outcome for HH, and at all 4 outcomes.

As you can see the lowest number of outcomes in the event is 0, when it does not occur, and the highest is equal to the all numbers of outcomes (outcomes in the sample space). So the probability will be always from $P(E)= \frac{0}{number~ of~ outcomes~ in ~ the ~ sample~ space}$ to $P(E)= \frac{number~ of~ outcomes~ in ~ the ~ sample~ space}{number~ of~ outcomes~ in ~ the ~ sample~ space}$, or, the same, from 0 to 1. So we can write constrained formula for every probability:

$0<= P(E) <=1.$

Moreover, the probability of the event concluded all cases is always equal 1. For example, we toss a coin and the probability of the event “H or T” is equal $P(H~ or ~ T)= \frac{1}{2} + \frac{1}{2}=1$. In the case of the two coins tossed instantly, we have $P(HH~ or ~ HT ~or ~ TH~ or ~ TT)= \frac{1}{4} + \frac{1}{4}+\frac{1}{4} + \frac{1}{4}=1$.

The Multiplication.

Now consider two events: we toss a coin and then toss a die. What is the probability for the outcome H on the coin and 5 on the die? To consider this task we need to multiply the probability of the first event by the probability of the second event. The probability of the event “H on the coin” is equal $\frac{1}{2}$, the probability of the event “5 on the die” is equal $\frac{1}{6}.$ So the asked probability is $P(E)=\frac{1}{2}*\frac{1}{6}= \frac{1}{12}.$

What with “T” on the coin and “2 or 3” on the die? $P(E)=\frac{number ~ that ~ Tails ~ occurs}{number~of ~ all ~ posible~ cases~for~ the ~ coin}* \\ * \frac{number ~ that ~ 2 or 3 ~ occurs}{number~of ~ all ~ posible~ cases~for ~ the ~ die} = \frac{1}{2}*\frac{2}{6}= \frac{2}{3}.$

Exclusive probability.

The probability for all cases is equal 1. If we know that there are a class of the 20 boys and the 10 girls, we choose randomly a person, what is the probability that it is a boy?
There are 30 persons at all and 20 boys, then asked probability is $P(B)=\frac{20}{30}=\frac{2}{3}.$ For a girl is $P(G)=\frac{10}{30}=\frac{1}{3}.$ As your can see for all possible cases, if we chose a boy or a girl $P(B~ or ~G)=\frac{2}{3}+\frac{1}{3}=1.$ So as we said total probability is always equal 1.

If we know that there are women and men in the city and we go to the street, the probability that we firstly meet a man is $\frac{2}{5}.$ What is the probability that we meet a woman firstly? We can use the rule of exclusive: the probability that it is a woman or a man is equal 1, the probability tat it is a man is $\frac{2}{5}$, and $P(M~ or ~W)= P(M)+P(W)=1.$ Then $P(W)= P(M)+P(M~or~W)=1 - \frac{2}{5} = \frac{3}{5}.$
So, the asked probability for a woman is equal $\frac{2}{5}.$

The same, if we know that there are in the parking either blue or green cars and no other colours and there will be a robbery. The probability that a hijacker get a blue car is $\frac{1}{20}.$ Then we can immediately say that probability that a green car will be stolen is $P(G)= P(G or B) - P(B)= 1 - \frac{1}{20} = \frac{19}{20}.$

Thus, if you see a confuse task like: “there are 254 tickets for a lottery, which were sold for \$2.5 each, while Jimmy bought 3 tickets and you bought 15. What is your chance to win (in the form of probability) if there are only five winning tickets”, now you should know the answer. At first, look at the question, then look at the information. We asked about the probability to win and have 5 winning tickets of 254, while we bought 15 of them. We do not care about Jimmy and the cost.

We know all we need: the number of outcomes in the sample space is 254 and the number of outcomes in the event if we bought one lottery is 5 (winning). So the probability for one bought ticket is $P(1 ~ ticket)$ is equal $\frac{5}{254},$ however we bought 15 of them, so we multiply by 15 and have $\frac{75}{254}.$

If you do not get some stuff, do not be upset because in the next post there will be the graphical explains and it will be easer to understand.