Basic Probability, part 5

August 24, 2009

Let us revise some information.

Pretend we at random choose two cards from a deck at the same time and we want to know what is the probability that we choose a queen or a king. As we know we these two events are mutually exclusive, because one card cannot be and a king and a queen at the same time, so we need to add two probabilities: $P(queen ~ or ~ king)=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}.$

Independent events.

And now consider the case when we toss a coin an the choose at random on card from a deck. What is the probability that we get Heads and an ace? As we know we need multiple probabilities of those two events: $P(H ~ or ~ Ace)=\frac{1}{2}*\frac{4}{52}=\frac{1}{26}.$ Current two events are called independent, because the second event “choose an ace” is not depend on the first event “get Heads”. The probability of getting an ace is $P(Ace)=\frac{4}{52}$ and it is independent weather we toss coin or not. The same for coin.

The multiplication rule 1:
If event A and event B are independent, then
$P(A~ and~ B)=P(A)*P(B).$

Dependent events.
Let us get a queen from an ordinary deck with 52 cards. The probability of this event is $P(Queen~ 1)=\frac{4}{52}.$ Now put the queen in a pocket. And try to get another queen from the deck. Notice that we have 51 cards and 3 queens! So the second event to get a queen has probability $P(Queen ~2)=\frac{3}{51}.$

These two events are called dependent events, the second event is dependent from the first. The probability that event B occurs given that event A has already occurred is denoted as $P(B~|~A).$

In our case with two queens the probability that we get a queen in the first choose and it is not replaced, and we get a queen in the second choose is:

$P(Queen ~1 ~ and~Queen~2)=P(Queen~1)*P(Queen~2|~Queen~1)=\\ = \frac{4}{52}*\frac{3}{51}=\frac{1}{221}.$

The multiplication rule 2:

If event A and event B are dependent, then
$P(A~ and~ B)=P(A)*P(B~|~A).$