# Dependent probability formula

G

#### Guest

##### Guest
What is the formula of a dependent probability for P(A and B), if it's different from the formula for a independent formula for P(A and B)?

Independent ---> P(A and B)= P(A) * P(B)

Dependent --->P(A and B)= ???????

#### soroban

##### Elite Member
Hello, Amy-Marie!

What is the formula of a dependent probability for P(A and B),
if it's different from the formula for a independent formula for P(A and B)?

Independent ---> P(A and B) = P(A) * P(B)

Dependent --->P(A and B)= ???
In symbols: $$\displaystyle \,P(A\,\cap\,B)\;=\;P(A)\cdot P(B|A)$$

In English: The probability that A happens . . . times . . .
$$\displaystyle \;\;$$the probabiity that B happens, given that A has already happened.

Example: Two cards are drawn from a deck of cards without replacement.
. . . . . . . . What is the probabiity that both are Hearts?

A = "first card is a Heart"
B = "second card is a Heart"

Since there are 52 cards of which 13 are Hearts: $$\displaystyle \,P(A)\,=\,\frac{13}{52}$$

What is $$\displaystyle P(B)$$ . . . prob. that the second is a Heart?
We might argue that it depends!
$$\displaystyle \;\;$$It depends on whether the first card was a Heart ... or not.

But the formula says that we can assume that the first card <u>was</u> a Heart.
$$\displaystyle \;\;$$So there are 51 cards left and 12 of them are Hearts: $$\displaystyle \;P(B)\,=\,\frac{12}{51}$$

Therefore: $$\displaystyle \,P(A\,\cap\,B)\:=\:\left(\frac{13}{52}\right)\left(\frac{12}{51}\right)\:=\:\frac{1}{17}$$

#### pka

##### Elite Member
While Soroban gave a really clear explanation in the case of clearly dependent events, that is clearly related events, you may not like to know that in general there is no formula for P(A and B).

Here I will write P(X and Y) as P(XY).

Say that events C & D are not independent.
Given P(C)=0.4 and P(D)=0.75, then there absolutely no way of finding P(CD) based on the given alone!

The best we can do is to give a set of bounds on P(CD).
We know that: $$\displaystyle 0.15 \le P(C \cap D) \le 0.4$$.

The first follows because the probability measure is monotone:
$$\displaystyle C \cap D \subseteq C\quad \Rightarrow \quad P(C \cap D) \le P(C)$$.

In general, $$\displaystyle 1 \ge P(C \cup D) = P(C) + P(D) - P(C \cap D)\quad \Rightarrow \quad P(C \cap D) \ge P(C) + P(D) - 1$$.