# probability P(A and B)

I have difficult to understand the following rule. Can anyone use a simple example to explain the rule to me? Thanks.

If A and B are any two events, then $P(A \cap B) = P(B)\cdot P(A| B)=P(A)\cdot P(B|A)$, where P(B|A) is the probability of B happening under the condition of event A.

• Shouldn't that be $P(B)P(A|B)$? Have you read en.wikipedia.org/wiki/Conditional_probability? Feb 28, 2017 at 4:10
• The first equality is certainly incorrect. Feb 28, 2017 at 4:20
• Included the missing $P(B)$ in $P(A\cap B)=\color{red}{P(B)}P(A\mid B)$. Be more careful with what you type. Feb 28, 2017 at 4:31
• @JMoravitz, Thanks. I just checked the book. there is no P(B). I think you are right, and the book is wrong on this part. Thank you very much!
– math
Feb 28, 2017 at 4:36
• Do you not undersatnd the correct equality? Or is the misundersanding purely due to the book's error?
– Kaz
Feb 28, 2017 at 14:35

Conditional probabilities are especially useful when it comes to things that involve having to pick things without replacement.

Consider a game of blackjack and you want to calculate the probability that you're dealt two aces. (We can assume for simplicity that we are only dealing with a single deck of 52 cards and that the first two cards in the deck are dealt to us). If we let $A$ be the event that the first card is an ace and let $B$ be the event that the second card is an ace, then it is clear that we want to find $P(A \text{ and } B)$.

Using the definition of conditional probability, we find, $$P(A \text{ and } B) = P(A)P(B|A) = P(\text{first card is an ace})P(\text{second card is an ace given the first card is an ace)}$$$$= \frac{4}{52}\cdot \frac{3}{51} = \frac{1}{221}.$$

We have that $P(A)$ is $\frac{4}{52}$ since there are 4 aces in a 52 card deck and $P(B|A) = \frac{3}{51}$ because after the first ace is drawn, there are 3 aces amongst the 51 cards remaining.

The Kolmogorov definition of conditional probability is $$P(B|A) = \frac{P(B \cap A)}{P(A)}$$ so rearranging gives $P(B|A) P(A) = P(B \cap A)$.

• This answers lacks the intuition OP appears to be looking for. Feb 28, 2017 at 19:42
• @Alexander The OP is not asking for intuition, but a simple explanation of why the equality holds. IMHO the best way to do this is from the definition. Feb 28, 2017 at 19:49
• He already has the def, he quotes it himself. "Can anyone use a simple example to explain the rule to me? " Feb 28, 2017 at 19:49

If $\mathbb{P}[A]=0$ then clearly we would like $\mathbb{P}[A \cap B]=0$ and this is the case.

Now if $\mathbb{P}[A]\neq0$ then this is equivalent to $$\mathbb{P}[B|A] = \frac{\mathbb{P}[A \cap B]}{\mathbb{P}[A]}$$ The right-hand side is a definition of $\mathbb{P}[B|A]$, that is, the probability of $B$ knowing $A$. It makes sense because it basically restricts the sample space to $A$.

Simple example: We throw a dice twice.

There are $36$ possible equiprobable outcomes, say $(a,b)$ with $1\leq a,b\leq 6$.

What is the probability of obtaining at least one $6$ knowing that the sum of the two throws is $8$ ?

Answer: There are $5$ outcomes for which the sum of the two throws is $8$. They are $$(2,6),(3,5),(4,4),(5,3),(6,2)$$ Of those $5$ outcomes, there are $2$ for which there is a $6$ : $(2,6)$ and $(6,2)$. Hence, reasonably, the probability should be $2/5$.

This example motivates the above definition.

Indeed, if you call $B$ the event "obtaining at least one $6$" and $A$ the event "the sum of the two throws is $8$" then we see that $$A \cap B = \{(2,6),(6,2)\}$$ and our answer $2/5$ can be written in the form $$\frac{2}{5} = \frac{2/36}{5/36} = \frac{\mathbb{P}[A\cap B]}{\mathbb{P}[A]}$$

Firstly, suppose that A and B are independent events, then the probability that they both occur is just the product of their individual probabilities: $P(A \cap B) = P(A)P(B)$.

Inthis situation, $P(A|B) = P(A)$ and $P(B|A) = P(B)$. This means that the probability of A occurring is the same whether or not B occurs, or vice versa.

Conditional probability lets us conceptualize dependent events. When there is some relationship between A and B, like A causes B or vice versa, or they have some common cause, then $P(A) \neq P(A|B)$ and $P(B) \neq P(B|A)$.

Example:

Suppose we have a jar of 100 marbles. 70 are red and 30 are green. Moreover, 65 have white stripes, and 35 are plain.

Thus: $P(R) = 70/100$, $P(G) = 30/100$, $P(S) = 65/100$ and $P(N) = 35/100$.

Now suppose there is a bias: suppose that most green marbles, 29 in fact, have a white stripe.

If we take all 30 green marbles out of the jar and put them into their own jar, the "green jar", and draw marbles from just the green jar, the probability is 29/30 that we get a striped one. This is obvious: we have 30 marbles, and 29 have stripes. This 29/30 probability is $P(S|G)$: probability of a stripe, when green is taken for granted.

Since 29 of the 65 striped marbles are green, it means that 36 are red. If we put all 70 red marbles into a jar, and choose from just that red jar, the probability of getting a striped marble is 36/70. This is $P(S|R)$.

Now we know that 29 green marbles have a stripe. So the probability of a striped green marble being chosen from the original red-green jar is precisely 29/100. That is $P(G \cap S)$.

If we plug in the numbers:

$$P(G \cap S) = P(S|G)P(G)$$

we see that it works out:

$$29/100 = 29/30 \times 30/100$$

The 30 and 30 cancel, leaving $29/10$.

It works with S and G reversed also. $P(G|S)$ is the probability of a marble being green, if it has a stripe. If we make a "striped jar" with just the striped marbles, it contains 65 marbles. 29 of them are a green. So $P(G|S)$ is $29/65$:

$$P(G \cap S) = P(G|S)P(S)$$

$$29/100 = 29/65 \times 65/100$$

Now the 65's cancel, leaving $29/100$.