Prove or disprove : $P[A|B] = P[B]$, the A and B are independent? Is this right? SOrry if this is extremely easy.
I did the following but I'm a little bit unsure about it:

Let $A=B$, and $P[A]>0$.
Then $$P[A|A] = P[A]$$
But A is not independent with itself:
$$P[AA] = P[A] \neq P[A]^{2} = P[A]P[A] $$
Thus, the preposition does not hold.

I'd appreciate your comments on this one.
Thanks!
 A: The error in the quoted reasoning is in  $P[A|A] = P[A]$. Instead we should have: $$P[A|A] = 1 \neq P[A]$$
For example, let the event be '$A = \text{it rains}$'.
Say you live in the desert where it almost never rains, but suppose that it is raining outside, then what is the probability that it is raining outside?

Edit: I notice now that you have to prove or disprove $P[A|B] = P[B]$. I misread this as $P[B|A] = P[B]$ which is true by definition when $A$ and $B$ are independent.

$P[A|B]=P[B]$ if the $A$ and $B$ are independent

This is not true. Counterexamples occur when $P[A] \neq P[B]$.
In case of independence you have by definition $P[A|B] = P[A]$ and that means that for any case where $P[A] \neq P[B]$ you have $$\rlap{\overbrace{\phantom{P[A|B] = P[A]}}^{\text{by definition}}} P[A|B] = \underbrace{P[A]\neq P[B]}_{\text{by assumption}}$$ Thus $P[A|B] = P[B]$ does not (necessarily) follow from independence, it is only true when also $P[A] = P[B]$ and the statement fails in cases where $P[A] \neq P[B]$.

If $P[A|B]=P[B]$, then the $A$ and $B$ are independent

This is neither true.
Counterexample: Any case where dependent $A$ and $B$ are dependent but $P[A|B]=P[B]$. (alternatively one can use cases with $P[A] \neq P[B]$)
An example is the case where we toss a fair coin two times and define $B$ be the event 'heads on the first toss', and let $A$ be the event 'heads on both tosses'. Then the table of probabilities looks as following
$$\begin{array}{ccc}
& \text{A: heads both tosses} & \text{not A: no heads both tosses} \\
\text{B: heads first toss} & 0.25 & 0.25 \\
\text{not B: no heads first toss} & 0 & 0.5 \\
\end{array}
$$
From these we can deduce $$P[A|B] = \frac{P[\text{$A$ and $B$}]}{P[\text{$A$ and $B$}]+P[\text{(not $A$) and $B$}]} = \frac{0.25}{0.25+0.25} = 0.5$$
and
$$P[B] = P[\text{$A$ and $B$}]+P[\text{(not $A$) and $B$}] = 0.25 + 0.25 = 0.5$$
And in this case we have $P[A|B]=P[B]$ but $A$ and $B$ are dependent. Therefore the statement is proven false by contradiction.
The condition $P[A|B]=P[B]$ does not (generally) imply that $A$ and $B$ are independent.
A: In order to avoid special cases let me consider that $P[A] \neq P[B]$ and both $<1$
I suggest you to think about conditioning as information gained.
So even intuitively $P[A|A]=P[A]$ is a nonsense. Indeed this is like to say "If I know that $A$ happen the probability that it happen remain unchanged". While it is obvious that $P[A|A]=1$
Now

Prove or disprove : $P[A|B]=P[B]$, the A and B are independent? Is
this right?

if we known that $P[A|B]=P[B]$ we can conclude that $A$ and $B$ are dependent events.
Indeed this is like to say "If we known that $B$ happen, then $P[A]$ change and become equal to $P[B]$.
A: To prove this first let's talk about the independent and not independent events:
Independent events is when we roll a die and then flip a coin. The number showing on the die has no effect on the coin that was tossed. Therefore these two events are independent.
Not independent would be the gender of each baby in a set of twins. If the twins are identical, then both of them will be male, or both of them would be female.
Since A and B are independent so:
= P(A) x P(B)
so
P[A | B] =  / P(B) = (P(A) x P(B)) / P(B) = P(A)

In other hand A is not independent with B:
 = P(A) x P(B|A)
P[A|B] =  / P(B) = P(A) x P(B|A) / P(B)

So A and B must be independent.
In case A and B independent and A = B:



