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I know that when two events $A$ and $B$ are independent then $P(A\mid B)=P(A)$, but still I did not find out examples of events where $P(A\mid B)\lt P(A)$ and $P(A\mid B)\gt P(A)$. I do not know whether it is possible or not, please any one help me to clarify this.

Clarifications:

  • $P(X)$ denotes the probability of event $X$.
  • $P(A\mid B)$ denotes the conditional probability of event $A$ given that event $B$ has already occurred.
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  • $\begingroup$ when you flip a coin, there is the possibility that it lands on its edge. $\endgroup$
    – JMP
    May 3 '15 at 9:14
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Let us now throw a die.

Take $A$ to be having an even outcome, take $B$ to be having at most $3$. Then $\textrm{Pr}(A\mid B)=1/3$ and $\textrm{Pr}(A)=1/2$.

Take $A$ to be having an odd outcome, take $B$ to be having at most $3$. Then $\textrm{Pr}(A\mid B)=2/3$ and $\textrm{Pr}(A)=1/2$.

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Suppose you roll a fair die.

Let $A$ be the event that you roll a six, and $B$ be the event that you roll a one.

Then $P(A|A)=1 > \frac{1}{6} = P(A)$, and $P(A|B)=0 < \frac{1}{6} = P(A)$.

In general it is about whether $B$ having happened makes $A$ more or less likely. Clearly if you know $A$ happens then $A$ will happen certainly and so has probability 1. Whereas if you know something that makes $A$ impossible happens then $A$ will not happen (and so has probability 0).

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2 blue and 3 red marbles are in a bag.

What are the chances of getting a blue marble?

The chance is 2 in 5

But after taking one out the chances change!

So the next time:

  • if we got a red marble before, then the chance of a blue marble next is 2 in 4
  • if we got a blue marble before, then the chance of a blue marble next is 1 in 4

In the above example A is the event of taking a blue marble out, and B is the previous event. As you see, different probabilities for B change the probability of A also.

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