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

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

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

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

In symbols: \(\displaystyle \,P(A\,\cap\,B)\;=\;P(A)\cdot P(B|A)\)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 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

\(\displaystyle \;\;\)It depends on whether the first card was a Heart ... or not.

But the formula says that we can

\(\displaystyle \;\;\)So there are 51 cards left and 12 of them are Hearts: \(\displaystyle \;P(B)\,=\,\frac{12}{51}\)

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

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

Given P(C)=0.4 and P(D)=0.75, then there absolutely no way of finding P(CD) based on the given alone!

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