Despite the value of knowing the probability of an event before it occurs, it can be even more valuable to know how learning part of the outcome changes the conditional probability.

## The Foundation for Understanding Conditional Probability

This article continues a series about probability, by introducing “conditional probability.”

If the terms are unfamiliar, consider reviewing “Introducing Probability Theory without Statistics“.

Recall that the “probability of an event” is a numeric value that states how likely it is for a specific outcome to occur, given the circumstances of an experiment in a given “probability space.” Simple examples include the fact that the probability of a tossed coin landing heads up is 0.5, and the probability of a six-sided dice roll showing ‘3’ is 1/6.

## Calculating Probability: Points to Consider

- The numeric value of a probability is any Real number from zero to one, inclusive. An event with “zero” probability cannot occur; if the probability is ‘1’ then the event must occur. As well, the probability that the experiment will yield some event in the probability space is exactly ‘1’.
- Mathematicians describe a event as “a set of outcomes”. For example, event ‘A’ might refer to a dice roll resulting in an odd number. Then “A = {1, 3, 5}” where the space of all outcomes is the set “S = {1, 2, 3, 4, 5, 6}”.
- Mathematically, the probability function of an event ‘A’ is shown as “P(A)”. A fair coin toss has “P(heads) = 0.5” and “P(tails) = 0.5”; a fair dice roll has “P(n) = 1/6 for n ∈ {1, 2, 3, 4, 5, 6}”.
- “Independent” events do not “depend on”, or influence, each other. As an example, I can toss a coin in Toronto without changing a poker hand being dealt in Las Vegas.

The probability that two events, ‘A’ and ‘B’, occur is the probability of the intersection set of their outcomes. This is shown as “P(A ∩ B)”. Consider a dice roll, where A = {1, 3, 5} = odd numbers, and B = {4, 5, 6} = high numbers. Then “A ∩ B = {5}”, and so “P(A ∩ B) = 1/6”.

The probability that two independent events will both occur equals the product of their separate probabilities. The equation P(A.B) = P(A)*P(B) expresses this, for a probability function ‘P’ governing events ‘A’ and ‘B’. Suppose the winning outcome is that you get heads from a coin toss, and then roll 1 or 2 using one die. The probability of these two independent events is one-half (the probability of getting heads in the coin toss) times one-third (the probability of rolling a one or a two on a six-sided die) which equals 1/6.

However, the events “odd” **and** “high” in one dice roll are * not* independent, since 0.5 * 0.5 = 0.25, but as we have seen, only {5} satisfies this condition with P({5}) = 1/6.

**Click to Read Page Two: Conditional Probability Calculations**

?? says

The probability of the sum of 2 die roles equaling 9, 10, 11, or 12 is 10 in 36, not 1 in 9.

Nimesh says

thanks