The set of all possible outcomes of an experiment is called the sample space.

For e.g. the sample space of outcomes from the roll of a single die is {1,2,3,4,5,6}.

The probability or likelihood of occurrence of an event is the number of instances of the sample space that satisfies the condition of interest.

For e.g. the probability of rolling an even number with a fair die is:

the number of even outcomes in the sample space {2,4,6} / total number of possible outcomes {1,2,3,4,5,6} = $1/2$ or 0.5

If P(A) is the probability of event A happening, then P(not A)=1-P(A)

Similarly, P(A ∪ B)=P(A)+P(B)-P(A ∩ B)

These two relationships are fairly evident when one draws a simple Venn diagram

Also, P(A ∩ B)=P(A)*P(B|A)=P(B)*P(A|B)

What this says is that the probability of A and B is the probability of event A times the probability of event B happening given A has happened.

This is useful when you have to find the probability given a condition that has already happened.

Let us look at some simple examples

e.g.1. an urn contains 5 white and 5 black marbles. two marbles are drawn from the urn, one at a time without replacement.

what is the probability that both marbles are black?

Well, the probability that the first marble is black is $5/10$ (P(A)). Since the marble is not replaced, the probability that the second marble is black is $4/9$ (P(B|A))

So the probability that both marbles are black=probability that the first marble is black AND the second marble is black (P(A ∩ B)), which is $5/10*4/9$=$2/9$

e.g.2. what is the probability in the above problem if the marble is replaced? Well, the probability that the first draw yields a black marble is $5/10$.

The probability that the second draw also yields a marble that is black is also $5/10$

So the probability that both the draws yield a marble that is black

= probability that the first draw yields a black marble AND the probability that the second draw yields a black marble=$5/10*5/10=1/4$

e.g.3. what is the probability in the above problem of there being at most one draw of a black marble, if sampling is always done with replacement?

Let's see. Probability that at most draw results in a black marble

= P(1st draw was white)*P(2nd draw was white)+P(1st draw was black)*P(2nd draw was white)+P(1st draw was white)*P(2nd draw was black)

=$5/10*5/10+5/10*5/10+5/10*5/10=0.75$

We can also calculate this another way. The probability that there was at most one black ball drawn = 1-P(both draws resulted in black ball draws)

Thus, our answer becomes: P(at most one black ball drawn)=1-P(both black balls drawn)=$1-5/10*5/10=0.75$

e.g.4. what is the probability of drawing a black card or a face card from a standard, well-shuffled deck of 52 playing cards?

There are 16 face cards in a standard deck. And 26 of the 52 cards are black. Of the 16 face cards, 8 are black.

So, if P(A)=probability of drawing a face card, and P(B)=probability of drawing a black card, then:

P(A ∪ B)=P(A)+P(B)-P(A ∩ B)=$16/52+26/52-8/52=34/52$

For e.g. the sample space of outcomes from the roll of a single die is {1,2,3,4,5,6}.

The probability or likelihood of occurrence of an event is the number of instances of the sample space that satisfies the condition of interest.

For e.g. the probability of rolling an even number with a fair die is:

the number of even outcomes in the sample space {2,4,6} / total number of possible outcomes {1,2,3,4,5,6} = $1/2$ or 0.5

If P(A) is the probability of event A happening, then P(not A)=1-P(A)

Similarly, P(A ∪ B)=P(A)+P(B)-P(A ∩ B)

These two relationships are fairly evident when one draws a simple Venn diagram

Also, P(A ∩ B)=P(A)*P(B|A)=P(B)*P(A|B)

What this says is that the probability of A and B is the probability of event A times the probability of event B happening given A has happened.

This is useful when you have to find the probability given a condition that has already happened.

Let us look at some simple examples

e.g.1. an urn contains 5 white and 5 black marbles. two marbles are drawn from the urn, one at a time without replacement.

what is the probability that both marbles are black?

Well, the probability that the first marble is black is $5/10$ (P(A)). Since the marble is not replaced, the probability that the second marble is black is $4/9$ (P(B|A))

So the probability that both marbles are black=probability that the first marble is black AND the second marble is black (P(A ∩ B)), which is $5/10*4/9$=$2/9$

e.g.2. what is the probability in the above problem if the marble is replaced? Well, the probability that the first draw yields a black marble is $5/10$.

The probability that the second draw also yields a marble that is black is also $5/10$

So the probability that both the draws yield a marble that is black

= probability that the first draw yields a black marble AND the probability that the second draw yields a black marble=$5/10*5/10=1/4$

e.g.3. what is the probability in the above problem of there being at most one draw of a black marble, if sampling is always done with replacement?

Let's see. Probability that at most draw results in a black marble

= P(1st draw was white)*P(2nd draw was white)+P(1st draw was black)*P(2nd draw was white)+P(1st draw was white)*P(2nd draw was black)

=$5/10*5/10+5/10*5/10+5/10*5/10=0.75$

We can also calculate this another way. The probability that there was at most one black ball drawn = 1-P(both draws resulted in black ball draws)

Thus, our answer becomes: P(at most one black ball drawn)=1-P(both black balls drawn)=$1-5/10*5/10=0.75$

e.g.4. what is the probability of drawing a black card or a face card from a standard, well-shuffled deck of 52 playing cards?

There are 16 face cards in a standard deck. And 26 of the 52 cards are black. Of the 16 face cards, 8 are black.

So, if P(A)=probability of drawing a face card, and P(B)=probability of drawing a black card, then:

P(A ∪ B)=P(A)+P(B)-P(A ∩ B)=$16/52+26/52-8/52=34/52$

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