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# Conditional probability

What’s the probability that event A occurs given that event B has already occurred?  That’s conditional probability.

Say that we are to find the chance that a randomly selected person is about to take a pill for headache. There would be a relatively “low” chance. But if we are then told that the selected person actually is having a headache, this would increase the chance for the person taking a pill.

## New info leads to different probability

Conditional probability is about revising a model after receiving new information will affect this model.

Say we are to find the chance of drawing a Queen from a standard deck of 52 cards. This probability is 4 out of 52 = 1/13. We are now given new information: A King has already been drawn without replacement from this deck of cards. The deck still has 4 Queens but now only 51 cards in total to draw from. This changes the probability from 4 out of 52 to 4 out of 51.

Let A be that you draw a Queen

Let B be that you draw a King

The chance of drawing a Queen given that a King has already been drawn, or in other words, the chance for both events to occurre, is reflected as the joint probability (ref Sample space, events & probabilities). P(A and B)  is 4/52 times 4/51 P(B): This formula essentially means that the chance of A given B equals the probability of A and B reduced to the sample space of B. Let’s plug our values into the formula and see if we get the result of the 4 out of 51 as we know it must be: The formula confirms that the probability of drawing a Queen given that a King has already been drawn without replacement is 4 out of 51. So, we confirm that P(A|B) = P(B).

## Probability of rolling odd number > 4?

What’s the chance of observing an odd number of a throw with a six-sided die? It’s 0.5. Now, what’s the chance of observing an odd number if we know that the throw is 4 or greater?

Let A be the event that you observe odd numbers and B be the event that the number is 4 or greater.

We would say what’s the probability of A given B = P(A|B) which, in this case, would be: What’s the probability of observing an odd number given that you have thrown 4 or greater?

From our initial sample space of 6 elements, we now have a reduced sample space of only three: {4,5,6}, and out of these three, we have one odd number which is 5. So, A given B is, in this case, 1/3.

## Learning resources on conditional probability #### Carsten Grube

Freelance Data Analyst

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