__
Probability__

If two events are **independent**,
the probability of one or the other (or both) happening is given by the formula

probability of event 1 + probability of
event 2 - the probability that both happen

What is the probability of getting a
king OR a spade in cards?

[p(king) = 4/52] + [p(spade) = 1/4] -
[4/52 x 1/4] = 4/13

**Tree diagrams**

-Most of the time there is no need to
draw a tree diagram

-When you use tree diagrams, you
multiply as you go along the branches

If there is a question asking for the
probability of getting at least one head there is no need to work out all the
different results that have "at least one head", simply work out the probability
of getting no heads at all and subtract this from 1!

**Conditional probability**

So far we have only looked at situations
when the events do not affect each other, however in the table above you can see
that, for example, if it is dry, it is twice as likely to be hot than if it is
wet so certain outcomes affect others.

What is the probability that it hot, **
given that it is dry**? (Answer: It is hot on 4 out of 5 dry days so 4/5)

**Note**

Tossing coins a number of times etc. is
NOT an example of conditional probability. The chance of tossing a head on the
next turn is totally unaffected by previous results. Even if you got six heads
in a row when tossing an unbiased coin, it does not make the next result any
more likely to be a tail!

**Examples of conditional probability**

**Given that **the first card I drew
was a king of spades, what is the probability that the next will be a king IF I
DO NOT REPLACE THE FIRST CARD.

P(king of spades) [which is 1
because I know it has happened] x P(king) [which is 3/51 because I am one king
down] = ^{3}/_{51}

However, if the question had asked for the
probability of a king of spades followed by any king, the answer is different
because it is not definite that I will get the king of spades.

P(king of spades) [which is 1/52] x
P(king) [which is 3/51 because I am one king down] = ^{1}/_{884}